1.6 Series, power series, and uniform convergence

An infinite sum is the limit of its partial sums. Series are tamed by comparison with the geometric series, and functions are tamed by uniform convergence, which lets limits pass through continuity, integrals and derivatives. Power series, e^x, sin and cos, and Euler's formula are the reward.

In school, $1 + \frac12 + \frac14 + \dots = 2$ was a fact about adding infinitely many things, and $e^x = 1 + x + \frac{x^2}{2!} + \dots$ was a formula to memorise. Adding infinitely many numbers makes sense only if we say what it means. The answer is a limit, so this chapter is 1.2 applied to partial sums. Then comes the harder idea: when the terms are functions, “the sum converges” can mean two different things, and only the stronger one lets continuity, integrals and derivatives pass through the sum. With that, power series become a tool you can trust, and Euler’s formula, promised in 0.6, gets its proof.

Before you start

Chapter 1.2: the $\varepsilon$ - $N$ game, monotone convergence, Cauchy sequences. Chapter 1.3: continuity with $\varepsilon$ and $\delta$ . Chapter 1.5: the integral and the fundamental theorem of calculus.

Quick checkA sequence of real numbers converges if and only if it is:

Quick checkAn increasing sequence that is bounded above:

Quick checkIf

|f(x) - g(x)| < \frac\varepsilon3

|g(x) - g(y)| < \frac\varepsilon3

and

|g(y) - f(y)| < \frac\varepsilon3

, what does the triangle inequality give for

|f(x) - f(y)|

The picture: two ways to converge

First, numbers. The partial sums of $\frac12 + \frac14 + \frac18 + \dots$ are $S_n = 1 - 2^{-n}$ : $0.5,\ 0.75,\ 0.875,\ \dots$ , each one closing half of the remaining gap to $1$ . The series “equals” $1$ because the partial sums converge to $1$ .

Now functions. Let $f_n(x) = x^n$ on $[0, 1]$ . At a fixed $x < 1$ the values $x^n$ shrink to $0$ ; at $x = 1$ they stay at $1$ .

$x$	$x^{10}$	$x^{100}$
$0.5$	$0.00098$	$\approx 0$
$0.9$	$0.349$	$0.0000266$
$0.99$	$0.904$	$0.366$
$1$	$1$	$1$

At every single point the sequence converges, but near $1$ you must go further and further out in $n$ to get small: the “bad region” slides toward $1$ and never disappears. So the graphs of $x^n$ never fit into a thin tube around the limit function ( $0$ for $x < 1$ , jumping to $1$ at $x = 1$ ). Each $f_n$ is continuous, yet the limit has a jump. This is the central warning of the chapter.

Where this came from· Oresme and the harmonic series

Nicole Oresme, a French scholar, showed around 1350 that $1 + \frac12 + \frac13 + \frac14 + \dots$ has no finite sum, even though its terms shrink to zero. He grouped the terms: $\frac13 + \frac14 \ge \frac12$ , then $\frac15 + \dots + \frac18 \ge \frac12$ , and so on, so there are infinitely many blocks each worth at least $\frac12$ . The argument was forgotten and found again in the 1600s (Pietro Mengoli in 1647, then the Bernoulli brothers). It set the pattern for this chapter: terms going to zero is necessary for a series to converge, but not sufficient.

The idea, in words

A series is a sequence in disguise: instead of the terms $a_k$ we look at the running totals $S_n = a_1 + \dots + a_n$ , and everything we proved about sequences in 1.2 applies to $(S_n)$ . Three questions organise the first half of the chapter. Does $(S_n)$ converge? (Tests.) Does it still converge if we take absolute values, so that no cancellation helps? (Absolute convergence, which is the safe kind.) And does the sum depend on the order of the terms? (Only for the unsafe kind.)

Then we let each term depend on $x$ . A series of functions defines a new function, and we want its properties: continuity, integral, derivative. These do not pass through a pointwise limit, as the $x^n$ picture shows. They do pass through a uniform limit, which means one $N$ works for all $x$ at once, the same $\delta$ -for-everywhere idea as uniform continuity in 1.3.

Why· The pattern in every proof here

Almost every proof in this chapter is one of three moves: compare with a geometric series (tests), apply the Cauchy criterion with the triangle inequality (absolute convergence, M-test), or split the error into three or two pieces and give each part of the budget $\varepsilon$ (uniform limits). Look for the move before reading each proof.

Series: definitions and first tests

Definition B

Series

Given a sequence $(a_k)$ , the series $\sum_{k=1}^\infty a_k$ converges to $S$ if the sequence of partial sums converges to $S$ ; then $S$ is its sum. If $(S_n)$ does not converge the series diverges.

In one lineAn infinite sum $\sum_{k=1}^\infty a_k$ , defined as the limit of the partial sums when that limit exists.

Why it existsInfinitely many additions cannot be performed, so the only honest meaning of an infinite sum is a limit of finite ones.

Definition B

Partial sum

The $n$ -th partial sum of $\sum a_k$ is $S_n = a_1 + a_2 + \dots + a_n$ . So $a_n = S_n - S_{n-1}$ for $n \ge 2$ .

In one line $S_n = a_1 + a_2 + \dots + a_n$ , the total of the first $n$ terms.

Why it existsEach partial sum is an ordinary finite sum, and the series is defined from them.

Theorem B Abbott Example 2.7.5

Geometric series

For $|r| < 1$ , $\displaystyle\sum_{k=0}^\infty r^k = \frac{1}{1 - r}$ . For $|r| \ge 1$ the series diverges.

In one line $\sum_{k=0}^\infty r^k = \dfrac{1}{1 - r}$ when $|r| < 1$ , and the series diverges when $|r| \ge 1$ .

Why it is trueThe partial sum has a closed form, and then the question is only whether r to the power n tends to zero.

Why· Scratch work: where the proof comes from

Multiply $S_n = 1 + r + \dots + r^n$ by $r$ and subtract: $S_n - rS_n = 1 - r^{n+1}$ , so $S_n = \frac{1 - r^{n+1}}{1 - r}$ (for $r \ne 1$ ). We want $S_n \to \frac1{1-r}$ , so the distance is $|S_n - \frac1{1-r}| = \frac{|r|^{n+1}}{|1 - r|}$ . Everything depends on showing $|r|^n \to 0$ for $|r| < 1$ , which needs an $\varepsilon$ - $N$ proof. Trick: write $|r| = \frac1{1 + h}$ with $h > 0$ and use Bernoulli’s inequality $(1 + h)^n \ge 1 + nh$ (induction), so $|r|^n \le \frac1{1 + nh} < \frac1{nh}$ , which is below $\varepsilon$ once $n > \frac1{h\varepsilon}$ .

ProofAbbott 2.7.5

The idea: Closed form for S_n; then |r|^n → 0 via Bernoulli's inequality. For |r| ≥ 1 the terms do not tend to 0.

For

r \ne 1

S_n = \dfrac{1 - r^{n+1}}{1 - r}

Multiply by

r

and subtract; all middle terms cancel.

Write

|r| = \dfrac1{1+h}

with

h > 0

0 < |r| < 1

then

\frac1{|r|} > 1

, so

\frac1{|r|} = 1 + h

for some

h > 0

.If

r = 0

then

r^n = 0

for

n \ge 1

, so there is nothing to prove.

|r|^n = \dfrac{1}{(1+h)^n} \le \dfrac{1}{1 + nh} < \dfrac{1}{nh}

Bernoulli:

(1 + h)^n \ge 1 + nh

, proved by induction since

(1+h)^{n+1} \ge (1 + nh)(1 + h) \ge 1 + (n+1)h

r^n \to 0

Choose

N \ge \frac1{h\varepsilon}

(Archimedean property). Then for

n > N

\frac1{nh} < \varepsilon

Hence

S_n \to \dfrac{1}{1-r}

Limit laws of 1.2:

S_n = \frac{1}{1-r} - \frac{r}{1-r}\,r^n

For

|r| \ge 1

the series diverges.

|r| \ge 1

then

|r^k| \ge 1

, so the terms do not tend to

0

, and a convergent series must have terms tending to

0

(nth-term test below).

\blacksquare

Worked example

Write

0.999\ldots

as a series and find its value.

Solution.

Series form. $0.999\ldots = \frac9{10} + \frac9{100} + \dots = 9\sum_{k=1}^\infty 10^{-k}$ . Each decimal digit is a term.
Pull out the first term. $\sum_{k=1}^\infty 10^{-k} = \frac1{10}\sum_{j=0}^\infty \left(\frac1{10}\right)^j = \frac1{10}\cdot\frac{1}{1 - \frac1{10}} = \frac19$ . Geometric series with $r = \frac1{10}$ , $|r| < 1$ .
Multiply. $9 \cdot \frac19 = 1$ . Arithmetic.

Answer. $0.999\ldots = 1$ exactly: the partial sums are $1 - 10^{-n}$ , which converge to $1$ . $\blacksquare$

Your answerFind

\displaystyle\sum_{k=1}^\infty 2\left(\frac35\right)^k

. Careful: the sum starts at

k = 1

Theorem B Abbott Thm 2.7.3

The nth-term test

If $\sum a_k$ converges, then $a_k \to 0$ . So if $a_k \not\to 0$ , the series diverges.

In one lineIf $\sum a_k$ converges, then $a_k \to 0$ .

Why it is trueA term is the difference of two consecutive partial sums, and both tend to the same limit.

ProofAbbott 2.7.3

The idea: a_n = S_n − S_{n−1}, and both partial sums tend to the same limit S.

a_n = S_n - S_{n-1}

for

n \ge 2

Definition of the partial sum.

By the difference law of 1.2,

a_n \to S - S = 0

\blacksquare

Both

(S_n)

and the shifted sequence

(S_{n-1})

converge to the sum

S

The converse is false, and the standard example is the harmonic series, which 1.2 left as a stretch problem (1.2, last problem).

Theorem B Oresme, c. 1350; Abbott Sec. 2.7

The harmonic series diverges

The partial sums $H_n = 1 + \frac12 + \dots + \frac1n$ are unbounded, so $\sum \frac1k$ diverges.

In one line $\sum_{k=1}^\infty \frac1k$ diverges, although $\frac1k \to 0$ .

Why it is trueBlocks of 2^m terms each add to at least 1/2, so the partial sums are unbounded.

ProofOresme; 1.2 stretch problem

The idea: Compare H_{2n} with H_n: the extra n terms are each at least 1/(2n), so they add at least 1/2.

H_{2n} - H_n = \dfrac1{n+1} + \dots + \dfrac1{2n}

The extra terms are

\frac1{n+1}, \dots, \frac1{2n}

, and there are

n

of them.

H_{2n} - H_n \ge n \cdot \dfrac1{2n} = \dfrac12

Each of the

n

terms is at least the smallest one,

\frac1{2n}

(H_n)

is unbounded, hence not convergent.

\blacksquare

Apply this with

n = 1, 2, 4, 8, \dots

H_{2^m} \ge 1 + \frac m2

, which exceeds any bound.

Quick checkSince

\frac1k \to 0

, does

\sum \frac1k

converge?

Theorem B Abbott Thm 2.7.4

Comparison test

Suppose $0 \le a_k \le b_k$ for all $k$ . If $\sum b_k$ converges, so does $\sum a_k$ . If $\sum a_k$ diverges, so does $\sum b_k$ .

In one lineIf $0 \le a_k \le b_k$ and $\sum b_k$ converges, then $\sum a_k$ converges; if $\sum a_k$ diverges, so does $\sum b_k$ .

Why it is trueNon-negative terms make the partial sums increase, so being bounded above is enough, and a bigger series bounds a smaller one.

ProofAbbott 2.7.4

The idea: Partial sums of a_k increase and are bounded by the sum of the b_k; monotone convergence.

The partial sums

A_n = a_1 + \dots + a_n

are increasing.

Each term is non-negative.

A_n \le B

for all

n

Compare termwise:

A_n \le B_n

, and

B_n \le \sum b_k = B

because the

B_n

increase to

B

(A_n)

converges. The second statement is the contrapositive.

\blacksquare

Monotone convergence theorem (1.2).

Worked example Abbott 2.7

Decide whether

\sum \frac1{k^2}

and

\sum \frac1{\sqrt k}

converge.

Solution.

First series: compare with a telescoping one. For $k \ge 2$ , $\frac1{k^2} \le \frac1{k(k-1)} = \frac1{k-1} - \frac1k$ . Making the denominator smaller makes the fraction bigger, and the partial fractions give a telescoping sum.
Sum the bound. $\sum_{k=2}^n \left(\frac1{k-1} - \frac1k\right) = 1 - \frac1n \le 1$ . All middle terms cancel.
Conclude. The partial sums of $\sum \frac1{k^2}$ are at most $1 + 1 = 2$ , so it converges (its value is $\frac{\pi^2}{6} = 1.6449\ldots$ , which the first ten terms approach as $1.5498$ ). Comparison test.
Second series. For $k \ge 1$ , $\frac1{\sqrt k} \ge \frac1k$ , and $\sum \frac1k$ diverges, so $\sum \frac1{\sqrt k}$ diverges. Comparison test, divergence form.

Answer. $\sum \frac1{k^2}$ converges; $\sum \frac1{\sqrt k}$ diverges. $\blacksquare$

Worked example

Find the exact sum

\displaystyle\sum_{k=1}^\infty \frac{1}{k(k+2)}

Solution.

Split into partial fractions. $\frac{1}{k(k+2)} = \frac12\left(\frac1k - \frac1{k+2}\right)$ . Check: $\frac1k - \frac1{k+2} = \frac{2}{k(k+2)}$ . The aim is to make neighbouring terms cancel.
Write the partial sum. $S_n = \frac12\left[\left(1 - \frac13\right) + \left(\frac12 - \frac14\right) + \left(\frac13 - \frac15\right) + \dots + \left(\frac1n - \frac1{n+2}\right)\right]$ . The partial sum is a finite sum, so we may regroup it freely.
Cancel. Every $\frac1j$ with $3 \le j \le n$ appears once with a plus and once with a minus, so $S_n = \frac12\left(1 + \frac12 - \frac1{n+1} - \frac1{n+2}\right)$ . Only the first two positive terms and the last two negative terms survive.
Take the limit. $\frac1{n+1}$ and $\frac1{n+2}$ tend to $0$ , so $S_n \to \frac12\cdot\frac32 = \frac34$ . Limit laws of 1.2.

Answer. $\frac34$ . (Check: the first four terms give $0.3333 + 0.125 + 0.0667 + 0.0417 = 0.5667$ and the sum creeps up to $0.75$ .) $\blacksquare$

Absolute convergence

Definition B

Absolute convergence

$\sum a_k$ converges absolutely if $\sum |a_k|$ converges. A series that converges but not absolutely converges conditionally.

In one line $\sum a_k$ converges absolutely if $\sum |a_k|$ converges.

Why it existsIt is the safe kind of convergence: no cancellation is needed, so the order of the terms cannot matter.

Theorem B Abbott Thm 2.7.2

Cauchy criterion for series

$\sum a_k$ converges iff $\forall \varepsilon > 0\ \exists N\ \forall n > m \ge N:\ |a_{m+1} + a_{m+2} + \dots + a_n| < \varepsilon$ .

In one line $\sum a_k$ converges iff for every $\varepsilon > 0$ there is $N$ with $|a_{m+1} + \dots + a_n| < \varepsilon$ for all $n > m \ge N$ .

Why it is trueIt is the Cauchy criterion of 1.2 applied to the partial sums, with S_n − S_m written as a block of terms.

ProofAbbott 2.7.2

The idea: S_n − S_m is exactly the block a_{m+1} + ... + a_n, so the condition says (S_n) is Cauchy; 1.2 says Cauchy is the same as convergent.

S_n - S_m = a_{m+1} + \dots + a_n

for

n > m

Definition of partial sums.

So the condition says

(S_n)

is a Cauchy sequence.

The condition is then

|S_n - S_m| < \varepsilon

for all

n > m \ge N

, the definition of Cauchy.

A real sequence is Cauchy iff it converges.

\blacksquare

Cauchy criterion for sequences (1.2).

Theorem B Abbott Thm 2.7.6

Absolute convergence implies convergence

If $\sum |a_k|$ converges, then $\sum a_k$ converges and $\left|\sum a_k\right| \le \sum |a_k|$ .

In one lineIf $\sum |a_k|$ converges, then $\sum a_k$ converges, and $\left|\sum a_k\right| \le \sum |a_k|$ .

Why it is trueCancellation can only make a block of terms smaller in size, so if the sizes are Cauchy, the signed terms are too.

Why· Scratch work: where the proof comes from

We have a hypothesis about $\sum |a_k|$ and a conclusion about $\sum a_k$ , and the Cauchy criterion for series is stated with blocks. So write the block for $a$ and compare it with the block for $|a|$ : by the triangle inequality $|a_{m+1} + \dots + a_n| \le |a_{m+1}| + \dots + |a_n|$ . The right side is small for large $m, n$ by the Cauchy criterion applied to the convergent series $\sum |a_k|$ (with the same $\varepsilon$ ). No splitting of the budget is needed: one comparison.

ProofAbbott 2.7.6

The idea: Triangle inequality turns the block of signed terms into a block of absolute values, which the Cauchy criterion makes small.

Let

\varepsilon > 0

. There is

N

with

|a_{m+1}| + \dots + |a_n| < \varepsilon

for all

n > m \ge N

Cauchy criterion for the convergent series

\sum |a_k|

For such

m, n

|a_{m+1} + \dots + a_n| \le |a_{m+1}| + \dots + |a_n| < \varepsilon

Triangle inequality, extended to many terms by induction.

\sum a_k

converges.

Cauchy criterion for series, in the other direction.

The inequality

\left|\sum a_k\right| \le \sum |a_k|

follows.

\blacksquare

\left|\sum_{k \le n} a_k\right| \le \sum_{k \le n} |a_k|

for each

n

; pass to the limit.

Proof workshopRebuild the proof that absolute convergence implies convergence

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick checkWhich series converges absolutely?

Theorem B Abbott Exercise 2.7.9; Rudin Thm 3.34

Ratio test

Let $a_k \ne 0$ and $\left|\dfrac{a_{k+1}}{a_k}\right| \to r$ . If $r < 1$ , $\sum a_k$ converges absolutely. If $r > 1$ , $\sum a_k$ diverges. If $r = 1$ , the test says nothing.

In one lineIf $\left|\frac{a_{k+1}}{a_k}\right| \to r$ , then $\sum a_k$ converges absolutely when $r < 1$ and diverges when $r > 1$ ; $r = 1$ decides nothing.

Why it is trueEventually the terms shrink at least as fast as a geometric series with ratio a little above r, which converges by comparison.

Why· Scratch work: where the proof comes from

For $r < 1$ the ratio is eventually near $r$ , so eventually the terms shrink by a factor close to $r$ each step, which is like a geometric series. We need a number $s$ with $r < s < 1$ to be a safe ratio bound: the ratio will eventually be below $s$ . The midpoint $s = \frac{r + 1}2$ works, because it is strictly between. In the $\varepsilon$ - $N$ language of 1.2, this is using the definition of the limit with $\varepsilon = s - r = \frac{1 - r}{2}$ . After the index $N$ where ratios are below $s$ , each step multiplies the size by less than $s$ : $|a_{N+j}| \le |a_N| s^j$ . The finitely many terms before $N$ do not affect convergence.

ProofRudin 3.34

The idea: Case r < 1: from some N on, |a_{k+1}| < s|a_k| with s = (1+r)/2 < 1, so |a_{N+j}| ≤ |a_N| s^j, and the geometric series with ratio s converges; compare. Case r > 1: eventually |a_{k+1}| > |a_k|, so the terms do not tend to 0.

Suppose

r < 1

and let

s = \frac{1 + r}{2}

, so

r < s < 1

. There is

N

with

\left|\frac{a_{k+1}}{a_k}\right| < s

for all

k \ge N

Definition of the limit

r

, with

\varepsilon = \frac{1 - r}{2}

(so that

r + \varepsilon = s

|a_{N+j}| \le |a_N|\,s^j

for all

j \ge 0

Induction on

j

: multiply the inequality

|a_{k+1}| < s|a_k|

j

times.

\sum_j |a_N| s^j

converges.

Geometric series with ratio

s

0 < s < 1

, converges (theorem above); a constant multiple

|a_N|

does not affect that.

\sum |a_k|

converges:

\sum a_k

converges absolutely.

Comparison test, and dropping or adding finitely many terms does not change convergence.

Then

|a_k| \ge |a_N| > 0

for all

k \ge N

, so

a_k \not\to 0

Now

r > 1

: with

\varepsilon = r - 1

the ratio is eventually greater than

1

, so

|a_{k+1}| > |a_k|

from some

N

on.

\sum a_k

diverges.

The nth-term test.

The test cannot decide when

r = 1

\blacksquare

For

r = 1

\sum \frac1k

diverges and

\sum \frac1{k^2}

converges, and both have ratios tending to

1

Worked example

Show that

\sum_{k=1}^\infty \frac{k}{2^k}

\sum_{k=0}^\infty \frac{3^k}{k!}

and

\sum_{k=1}^\infty \frac{2^k k!}{k^k}

converge.

Solution.

First series: ratio. $\left|\frac{a_{k+1}}{a_k}\right| = \frac{(k+1)/2^{k+1}}{k/2^k} = \frac{k+1}{2k} \to \frac12$ . Divide $(k+1)$ by $k$ and $2^k$ by $2^{k+1}$ .
Decide. $\frac12 < 1$ , so it converges absolutely. Ratio test, $r < 1$ .
Second series: ratio. $\frac{3^{k+1}/(k+1)!}{3^k/k!} = \frac{3}{k+1} \to 0 < 1$ . $(k+1)! = (k+1)\,k!$ .
Decide. It converges (its sum is $e^3 = 20.0855\ldots$ , which the first 30 terms already give to ten places). Ratio test.
Third series: ratio. $\dfrac{2^{k+1}(k+1)!/(k+1)^{k+1}}{2^k k!/k^k} = 2\cdot\dfrac{(k+1)\,k^k}{(k+1)^{k+1}} = 2\left(\dfrac{k}{k+1}\right)^k = 2\left(1 + \dfrac1k\right)^{-k}$ . Cancel $2^k$ , $k!$ and one factor $(k+1)$ from $(k+1)^{k+1}$ .
Decide. $\left(1 + \frac1k\right)^k \to e$ , so the ratio tends to $\frac2e = 0.7358 < 1$ , and the series converges absolutely. Ratio test, $r < 1$ . The limit $\left(1 + \frac1k\right)^k \to e$ is the compound-interest limit used in the applications below.

Answer. All three converge. $\blacksquare$ (Later: $\sum \frac{k}{2^k} = 2$ , a term-by-term differentiation example.)

Quick checkFor

\sum \frac1{k^2}

the ratio

\left|\frac{a_{k+1}}{a_k}\right| = \frac{k^2}{(k+1)^2}

tends to

1

. What does the ratio test tell you?

Theorem B Abbott Thm 2.7.7

Alternating series test

If $b_1 \ge b_2 \ge \dots \ge 0$ and $b_k \to 0$ , then $\sum_{k=1}^\infty (-1)^{k+1} b_k = b_1 - b_2 + b_3 - \dots$ converges, and $|S - S_n| \le b_{n+1}$ .

In one lineIf $b_k$ decreases to $0$ , then $\sum (-1)^{k+1} b_k$ converges, and the error after $n$ terms is at most $b_{n+1}$ .

Why it is trueThe partial sums with an even number of terms go up, those with an odd number go down, and the gap between the two kinds is b, which shrinks to zero.

ProofAbbott 2.7.7

The idea: Even partial sums increase and are bounded above by S_1; odd partial sums are the even ones plus b_{2n+1} → 0, so they converge to the same limit.

The even partial sums increase:

S_{2n+2} \ge S_{2n}

Pair terms:

S_{2n+2} = S_{2n} + (b_{2n+1} - b_{2n+2})

and the bracket is

\ge 0

since

b

decreases.

They are bounded above by

S_1 = b_1

S_{2n} = b_1 - (b_2 - b_3) - \dots - (b_{2n-2} - b_{2n-1}) - b_{2n} \le b_1

, each bracket being non-negative.

S_{2n} \to L

for some

L

Monotone convergence theorem (1.2).

S_{2n+1} \to L

too.

S_{2n+1} = S_{2n} + b_{2n+1}

, and

b_{2n+1} \to 0

; limit laws.

The odd partial sums decrease:

S_{2n+1} \le S_{2n-1}

. So

S_{2n} \le L \le S_{2n+1}

for every

n

: an increasing sequence stays below its limit and a decreasing one stays above it.

S_{2n+1} = S_{2n-1} - (b_{2n} - b_{2n+1})

and the bracket is

\ge 0

. This is needed for the error bound at the end.

Hence

S_n \to L

If both the even-indexed and odd-indexed terms of a sequence tend to

L

, the whole sequence does: given

\varepsilon

, take the larger of the two starting points.

The error satisfies

|L - S_n| \le b_{n+1}

S_{2n}

increases to

L

and

S_{2n+1}

decreases to

L

, so

L

lies between any

S_n

and

S_{n+1}

, which differ by

b_{n+1}

\blacksquare

Worked example Abbott 2.7

The alternating harmonic series: show

1 - \frac12 + \frac13 - \dots

converges, but not absolutely, and bound the error of

S_{10}

Solution.

Converges. $b_k = \frac1k$ decreases to $0$ , so the alternating series test applies. Both hypotheses hold.
Not absolutely. $\sum |a_k| = \sum \frac1k$ is the harmonic series, which diverges. So the convergence is conditional.
Error bound. $S_{10} = 0.6456\ldots$ and $|S - S_{10}| \le b_{11} = \frac1{11} = 0.0909$ . Error of at most the first omitted term.

Answer. Conditionally convergent. Its sum is $\ln 2 = 0.6931\ldots$ (a Taylor series fact); the error of $S_{10}$ is $0.0475$ , within the bound. $\blacksquare$

Your answerFor

1 - \frac12 + \frac13 - \dots

, the error after

n

terms is at most

\frac1{n+1}

. What is the smallest

n

that guarantees an error below

0.001

Worked example

Classify each series as absolutely convergent, conditionally convergent or divergent:

\sum \frac{k}{k+1}

\sum \frac{(-1)^k}{k}

\sum \frac{(-1)^k}{k^2}

\sum (-1)^k

Solution.

Test the terms first: $\sum \frac{k}{k+1}$ . The terms tend to $1$ , not $0$ . If the terms do not tend to $0$ , the series diverges (nth-term test). Always do this check first because it is one line.
The last series, $\sum (-1)^k$ , fails the same test. Its terms are $\pm 1$ and do not tend to $0$ . The partial sums alternate between $-1$ and $0$ , which agrees: no limit.
Absolute values for $\sum \frac{(-1)^k}{k^2}$ . $\sum \left|\frac{(-1)^k}{k^2}\right| = \sum \frac1{k^2}$ , which converges. Comparison with the telescoping series, done earlier. Absolute convergence implies convergence.
Absolute values for $\sum \frac{(-1)^k}{k}$ . $\sum \frac1k$ diverges, so it is not absolutely convergent. Harmonic series.
Does it converge anyway? $\frac1k$ decreases to $0$ , so the alternating series test says yes. Both hypotheses of the test hold.

Answer. $\sum \frac{k}{k+1}$ diverges; $\sum \frac{(-1)^k}{k}$ converges conditionally; $\sum \frac{(-1)^k}{k^2}$ converges absolutely; $\sum (-1)^k$ diverges. $\blacksquare$

Rearrangements

Theorem B Abbott Sec. 2.7; Rudin Thm 3.55

Rearranging an absolutely convergent series

Let $\sum a_k$ converge absolutely with sum $S$ , and let $\sigma : \mathbb N \to \mathbb N$ be a bijection (a reordering). Then $\sum a_{\sigma(j)}$ converges to $S$ .

In one lineIf $\sum a_k$ converges absolutely, every rearrangement converges to the same sum.

Why it is trueOnly the first few terms matter for the sum up to ε; the tail is tiny in absolute value, so wherever the tail terms end up, they cannot change the total by more than ε.

ProofRudin 3.55

The idea: Choose N so the tail of absolute values after N is below ε. Wait until the rearranged series has used all of a_1..a_N; after that, its partial sums differ from S_N only by tail terms.

Let

\varepsilon > 0

. Choose

N

with

\sum_{k > N} |a_k| < \varepsilon

and

|S - S_N| < \varepsilon

Absolute convergence: the tails of

\sum |a_k|

go to

0

. Convergence of

\sum a_k

S_N \to S

. Take

N

large enough for both.

Choose

M

such that

\{1, \dots, N\} \subseteq \{\sigma(1), \dots, \sigma(M)\}

\sigma

is onto, so each of

1, \dots, N

equals

\sigma(j)

for some

j

; take

M

larger than all these

j

T_n - S_N

is a sum of distinct terms

a_k

with

k > N

For

n \ge M

, the rearranged partial sum

T_n = \sum_{j \le n} a_{\sigma(j)}

contains

a_1, \dots, a_N

once each, plus terms with index greater than

N

(each at most once, since

\sigma

is one-to-one).

|T_n - S_N| \le \sum_{k > N} |a_k| < \varepsilon

Triangle inequality, then the tail bound.

|T_n - S| \le |T_n - S_N| + |S_N - S| < 2\varepsilon

for all

n \ge M

\blacksquare

Triangle inequality through

S_N

Beyond the course: for understanding· Riemann's rearrangement theorem (stated, label B)

For conditionally convergent series the order matters completely. Riemann’s rearrangement theorem: if $\sum a_k$ converges conditionally, then for every real number $T$ (and also $\pm\infty$ ) there is a rearrangement whose sum is $T$ . The idea: the positive terms alone add to $+\infty$ and the negative terms to $-\infty$ (otherwise the series would converge absolutely or diverge). Take positive terms until the running total passes $T$ , then negative terms until it falls below $T$ , and repeat; since the terms tend to $0$ , the overshoots shrink and the total converges to $T$ .

Example with the alternating harmonic series, whose sum is $\ln 2 = 0.693\ldots$ : take one positive term, then two negative terms: $1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \dots$ . The sum becomes $\frac12\ln 2 = 0.3466\ldots$ (numerically, the first 300000 terms give $0.346572$ ). The same terms in a different order give half the answer.

Watch out· Why this matters

An expression like $\sum_{j,k} a_{jk}$ over a grid, or a product of two series, involves a choice of order. Absolute convergence is the condition that makes every order give the same answer, and the product of series below relies on it.

Quick checkWhich series can be added in any order without changing its sum?

Sequences of functions: pointwise versus uniform

From here on $D$ is a set of real numbers and $f_n : D \to \mathbb R$ are functions.

Definition B

Pointwise convergence

$f_n \to f$ pointwise on $D$ if

\forall x \in D\ \ \forall \varepsilon > 0\ \ \exists N\ \ \forall n > N:\quad |f_n(x) - f(x)| < \varepsilon.

Here $N$ may depend on both $\varepsilon$ and $x$ .

In one line $f_n \to f$ pointwise on $D$ if for every $x \in D$ the numbers $f_n(x)$ converge to $f(x)$ .

Why it existsIt is the weakest, most natural notion: apply sequence convergence at each point separately.

Definition B

Uniform convergence

$f_n \to f$ uniformly on $D$ if

\forall \varepsilon > 0\ \ \exists N\ \ \forall n > N\ \ \forall x \in D:\quad |f_n(x) - f(x)| < \varepsilon.

Compared with pointwise convergence, the quantifier $\forall x$ moved to the end, so $N$ is chosen before $x$ and cannot depend on it. Equivalently, $\sup_{x \in D} |f_n(x) - f(x)| \to 0$ .

In one line $f_n \to f$ uniformly on $D$ if for every $\varepsilon > 0$ there is ONE $N$ that works for all $x \in D$ at once.

Why it existsIt is exactly what makes the graphs of f_n fit inside a thin tube around f, and it is the condition under which limits can be exchanged with continuity, integrals and derivatives.

Why· The same quantifier lesson as uniform continuity

In 1.3, moving “for all points” from before $\delta$ to after $\delta$ turned continuity into uniform continuity. Here the same move turns pointwise into uniform convergence. In both cases the stronger notion is the one theorems need, because it makes a single tolerance control the whole domain.

Worked example Abbott Sec. 6.2

Show

x^n \to 0

pointwise on

[0, 1)

but NOT uniformly, and uniformly on

[0, \tfrac12]

Solution.

Pointwise. For fixed $x \in [0, 1)$ , $x^n \to 0$ (the geometric argument above). Each point is a separate sequence.
Not uniform: negation. Uniform convergence to $0$ fails if $\exists \varepsilon\ \forall N\ \exists n > N\ \exists x:\ |x^n| \ge \varepsilon$ . Negation rules of 0.1, applied to the definition.
Choose $\varepsilon$ and the bad points. Take $\varepsilon = \frac12$ . For each $n$ let $x_n = \left(\frac12\right)^{1/n} \in (0, 1)$ . Then $x_n^n = \frac12 \ge \varepsilon$ . We chose $x_n$ so that the $n$ -th function equals exactly $\frac12$ there: the bad point slides to $1$ as $n$ grows. For $n = 10$ , $x_{10} = 0.933$ .
Uniform on $[0, \frac12]$ . For $x \in [0, \frac12]$ , $|x^n| \le 2^{-n}$ , and $2^{-n} < \varepsilon$ once $n > \log_2\frac1\varepsilon$ . This $N$ does not involve $x$ . The bound is the same for every $x$ in the domain, which is exactly what ‘one $N$ for all $x$ ’ needs.

Answer. Pointwise but not uniform on $[0, 1)$ ; uniform on $[0, \frac12]$ . $\blacksquare$

Worked example

Decide whether

f_n(x) = \dfrac{x^2}{n}

converges uniformly on

[0, 3]

, and on

\mathbb R

Solution.

Find the pointwise limit. For each fixed $x$ , $\frac{x^2}{n} \to 0$ . So $f = 0$ . $x^2$ is a fixed number and $n \to \infty$ .
Compute the distance to the limit, at its worst point. $\sup_x |f_n(x) - f(x)| = \sup_x \frac{x^2}{n}$ . Uniform convergence means exactly that this supremum tends to $0$ .
On $[0, 3]$ . The largest value of $x^2$ is $9$ , so the supremum is $\frac9n \to 0$ . The bound $\frac{9}{n}$ does not depend on $x$ , so one $N$ serves all points: $N > \frac9\varepsilon$ .
On $\mathbb R$ . $x^2$ is unbounded, so $\sup_x \frac{x^2}{n} = \infty$ for every $n$ . In particular at $x = n$ the distance is $n$ , which does not tend to $0$ . There is always a point $x$ , depending on $n$ , where $f_n$ is far from $0$ . The bad point runs off to infinity.

Answer. Uniform on $[0, 3]$ , not uniform on $\mathbb R$ (though it converges pointwise on $\mathbb R$ ). $\blacksquare$

Your answerOn

[0, 0.9]

the functions

x^n

converge to

0

. What is

\sup_{[0, 0.9]} |x^n|

for

n = 10

? (Give a decimal.)

Quick checkWhich phrase describes uniform (not just pointwise) convergence?

Theorem B Abbott Thm 6.2.6

Uniform limit of continuous functions is continuous

If $f_n \to f$ uniformly on $D$ and each $f_n$ is continuous at $c \in D$ , then $f$ is continuous at $c$ .

In one lineIf each $f_n$ is continuous and $f_n \to f$ uniformly, then $f$ is continuous.

Why it is trueCompare f(x) with f(y) through f_n at both points: two uniform errors and one continuity error, each given a third of the budget.

Why· Scratch work: the epsilon over three split

We must show $|f(x) - f(c)| < \varepsilon$ for $x$ near $c$ . We know three things: $f_n$ is close to $f$ at $x$ (uniform convergence), $f_n$ is close to $f$ at $c$ (same), and $f_n$ is continuous at $c$ , so $f_n(x)$ is close to $f_n(c)$ . Go from $f(x)$ to $f_n(x)$ to $f_n(c)$ to $f(c)$ : three steps, each of size less than $\frac\varepsilon3$ , and the triangle inequality gives $<\varepsilon$ in total. The order of choices matters. First fix $\varepsilon$ . Then use uniform convergence to fix ONE function $f_n$ that is within $\frac\varepsilon3$ of $f$ everywhere. Only then use the continuity of that one $f_n$ to find $\delta$ . Uniformity is what lets us choose $n$ before $x$ : with merely pointwise convergence, the $n$ that works at $x$ differs from the one that works at $c$ , and the chain breaks.

ProofAbbott 6.2.6

Let

\varepsilon > 0

Arbitrary tolerance.

There is

n

such that

|f_n(t) - f(t)| < \frac\varepsilon3

for ALL

t \in D

Uniform convergence with tolerance

\frac\varepsilon3

: one

N

for all

x

. Pick any

n > N

.This is where uniformity is used: n is chosen before x.

There is

\delta > 0

with

|f_n(x) - f_n(c)| < \frac\varepsilon3

whenever

|x - c| < \delta

Continuity of the single function

f_n

c

, with tolerance

\frac\varepsilon3

For

|x - c| < \delta

|f(x) - f(c)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)| < \dfrac\varepsilon3 + \dfrac\varepsilon3 + \dfrac\varepsilon3 = \varepsilon

Triangle inequality along

f(x) \to f_n(x) \to f_n(c) \to f(c)

f

is continuous at

c

Definition of continuity at

c

\blacksquare

Proof workshopGive the reason for each line: uniform limits of continuous functions

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $\varepsilon > 0$ .
Choose $n$ with $|f_n(t) - f(t)| < \frac\varepsilon3$ for every $t \in D$ .
Choose $\delta$ with $|f_n(x) - f_n(c)| < \frac\varepsilon3$ when $|x - c| < \delta$ .
$|f(x) - f(c)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)|$
$< \frac\varepsilon3 + \frac\varepsilon3 + \frac\varepsilon3 = \varepsilon$ .

Where this came from· Cauchy's mistake, Abel's counterexample, Weierstrass's repair

In his Cours d’analyse (1821) Cauchy stated that a convergent series of continuous functions has a continuous sum. In 1826 Niels Henrik Abel pointed out that this is false, using the series $\sin x - \frac12\sin 2x + \frac13\sin 3x - \dots$ , which converges for every $x$ to a function with a jump at odd multiples of $\pi$ . The statement becomes true when “convergent” is strengthened to “uniformly convergent”. Philipp Seidel and George Stokes noticed the missing condition around 1847, and Karl Weierstrass made the idea precise and used it systematically; the M-test below carries his name. Episodes like this one are a large part of why analysis was rewritten in the careful $\varepsilon$ style taught today.

Worked example Abbott 6.2

Why can't

x^n

[0, 1]

converge uniformly? A second proof.

Solution.

The limit. $x^n \to f$ pointwise on $[0, 1]$ where $f(x) = 0$ for $x < 1$ and $f(1) = 1$ . The table of the picture.
Each $f_n$ is continuous, but $f$ has a jump at $1$ . So $f$ is not continuous. The values just left of $1$ are $0$ and the value at $1$ is $1$ .
Apply the theorem. If the convergence were uniform, $f$ would be continuous. It is not. Contrapositive of the uniform limit theorem.

Answer. The convergence is not uniform on $[0, 1]$ . $\blacksquare$ The uniform limit theorem is a way to detect failure of uniformity without finding the bad points.

Theorem B Abbott Thm 6.4.5

Weierstrass M-test

Let $f_k : D \to \mathbb R$ with $|f_k(x)| \le M_k$ for all $x \in D$ , where $\sum M_k < \infty$ . Then $\sum f_k$ converges uniformly on $D$ .

In one lineIf $|f_k(x)| \le M_k$ for all $x \in D$ and $\sum M_k$ converges, then $\sum f_k$ converges uniformly (and absolutely) on $D$ .

Why it is trueThe numbers M_k bound all the functions at once, so the tail of the series is small by the same amount at every x.

ProofAbbott 6.4.5

The idea: At each x the series converges absolutely by comparison, which defines the limit f(x). The distance from f(x) to the n-th partial sum is at most the tail of ΣM_k, a number that does not depend on x.

For each

x \in D

\sum |f_k(x)|

converges, so

f(x) = \sum f_k(x)

exists.

Comparison test with

M_k

, at each fixed

x

; absolute convergence implies convergence.

\left|f(x) - \sum_{k=1}^n f_k(x)\right| = \left|\sum_{k > n} f_k(x)\right| \le \sum_{k > n} M_k

Triangle inequality for the tail, then

|f_k(x)| \le M_k

termwise.

Given

\varepsilon > 0

choose

N

with

\sum_{k > n} M_k < \varepsilon

for all

n > N

The tail of a convergent series tends to

0

, and the bound does not contain

x

.This is the uniformity.

\sum f_k \to f

uniformly on

D

Then the distance is below

\varepsilon

for every

x

at once.

\blacksquare

Worked example Abbott Sec. 6.4

Show that

f(x) = \sum_{k=1}^\infty \frac{\sin kx}{k^2}

is defined and continuous on all of

\mathbb R

Solution.

Bound each term. $\left|\frac{\sin kx}{k^2}\right| \le \frac1{k^2} = M_k$ for all $x$ . $|\sin| \le 1$ .
The bounds are summable. $\sum \frac1{k^2}$ converges (comparison example above). This is the hypothesis of the M-test.
Apply the M-test. The series converges uniformly on $\mathbb R$ . Weierstrass M-test.
Each partial sum is continuous (a finite sum of sines), so the uniform limit $f$ is continuous. Uniform limit of continuous functions.

Answer. $f$ is a continuous function on $\mathbb R$ . $\blacksquare$

Quick checkWhy does the M-test give UNIFORM convergence and not just pointwise?

Power series

Definition B

Power series

A power series (centred at $0$ ) is an expression $\sum_{k=0}^\infty c_k x^k$ with real coefficients $c_k$ . For each $x$ it is an ordinary series, which may converge or diverge.

In one lineA series $\sum_{k=0}^\infty c_k x^k$ whose terms are constants times powers of $x$ ; a polynomial of infinite degree.

Why it existsPower series are the functions you can differentiate and integrate by rules, so they are the working tool for exp, sin, cos and for solving differential equations.

Lemma B Abbott Thm 6.5.1

Convergence of power series (Abel's lemma)

If $\sum c_k x_0^k$ converges, then $\sum c_k x^k$ converges absolutely for all $x$ with $|x| < |x_0|$ .

In one lineIf $\sum c_k x_0^k$ converges, then $\sum c_k x^k$ converges absolutely for every $|x| < |x_0|$ .

Why it is trueIf the series converges at x_0, its terms are bounded there, and then the terms at smaller x are dominated by a geometric series.

ProofAbbott 6.5.1

The idea: Terms tend to 0 at x_0, so they are bounded by some B; at x the terms are at most B(|x|/|x_0|)^k, a convergent geometric series.

c_k x_0^k \to 0

, so

|c_k x_0^k| \le B

for some

B

and all

k

The nth-term test: a convergent series has terms tending to

0

; a convergent sequence is bounded (1.2).

For

|x| < |x_0|

|c_k x^k| = |c_k x_0^k|\left|\dfrac{x}{x_0}\right|^k \le B\left|\dfrac{x}{x_0}\right|^k

Write

x^k = x_0^k (x/x_0)^k

\sum |c_k x^k|

converges.

Geometric series with ratio

\left|\frac{x}{x_0}\right| < 1

, then the comparison test.

\blacksquare

Definition B

Radius of convergence

The radius of convergence of $\sum c_k x^k$ is $R = \sup\{|x| : \sum c_k x^k \text{ converges}\} \in [0, \infty]$ . By Abel’s lemma, the series converges absolutely for $|x| < R$ (some convergence point $x_0$ has $|x_0| > |x|$ , by the approximation property of the supremum), and by definition it diverges for $|x| > R$ . At $x = \pm R$ anything can happen.

In one line $R = \sup\{|x| : \sum c_k x^k \text{ converges}\}$ : the series converges absolutely for $|x| < R$ and diverges for $|x| > R$ .

Why it existsIt says a power series converges on a symmetric interval around its centre, and tells you how wide, so you know where the function is defined.

Theorem B Abbott Exercise 6.5; Rudin Thm 3.39

Ratio formula for the radius

Suppose $c_k \ne 0$ and $\left|\dfrac{c_{k+1}}{c_k}\right| \to L \in [0, \infty]$ . Then $R = \dfrac1L$ .

In one lineIf $c_k \ne 0$ and $\left|\frac{c_{k+1}}{c_k}\right| \to L$ , then $R = \frac1L$ (with $R = \infty$ if $L = 0$ and $R = 0$ if $L = \infty$ ).

Why it is trueApply the ratio test to the series at a fixed x: the ratio of terms is |x| times the ratio of coefficients.

ProofRudin 3.39

The idea: At a fixed x ≠ 0 the ratio of consecutive terms tends to L|x|. The ratio test says converge if L|x| < 1 and diverge if L|x| > 1; that is |x| < 1/L and |x| > 1/L.

For

x \ne 0

\left|\dfrac{c_{k+1}x^{k+1}}{c_k x^k}\right| = \left|\dfrac{c_{k+1}}{c_k}\right||x| \to L|x|

Divide the

(k+1)

-st term by the

k

-th.

L|x| < 1

, i.e.

|x| < \frac1L

, the series converges absolutely.

Ratio test, case

r < 1

L|x| > 1

, i.e.

|x| > \frac1L

, the series diverges.

Ratio test, case

r > 1

Therefore

R = \frac1L

So the set of convergence points contains all

|x| < \frac1L

and none with

|x| > \frac1L

, whose supremum of

|x|

\frac1L

. If

L = 0

every

x

gives ratio

0 < 1

, so

R = \infty

; if

L = \infty

every

x \ne 0

gives ratio

\infty > 1

, so

R = 0

\blacksquare

Worked example

Find the radius and the set of convergence for

\sum \frac{x^k}{k\,3^k}

, and the radius for

\sum x^k

\sum \frac{x^k}{k!}

\sum k!\,x^k

Solution.

Ratio of coefficients. For $c_k = \frac1{k3^k}$ : $\left|\frac{c_{k+1}}{c_k}\right| = \frac{k\cdot 3^k}{(k+1)3^{k+1}} = \frac13\cdot\frac{k}{k+1} \to \frac13$ . Cancel $3^k$ and simplify.
Radius. $R = 3$ . Ratio formula, $L = \frac13$ .
The endpoints. At $x = 3$ the series is $\sum \frac1k$ , which diverges. At $x = -3$ it is $\sum \frac{(-1)^k}{k}$ , which converges by the alternating series test. The ratio test gave no information at $|x| = R$ , so test each endpoint separately.
The others. $\sum x^k$ has $L = 1$ , $R = 1$ . $\sum \frac{x^k}{k!}$ has ratio $\frac1{k+1} \to 0$ , so $R = \infty$ . $\sum k!\,x^k$ has ratio $k + 1 \to \infty$ , so $R = 0$ (it converges only at $x = 0$ ). Ratio formula in each case.

Answer. The first series converges exactly for $-3 \le x < 3$ . $\blacksquare$

Your answerFind the radius of convergence of

\displaystyle\sum_{k=0}^\infty \frac{4^k}{k+1}\,x^k

Quick checkA power series has radius of convergence

R = 2

. What can you conclude at

x = 2

Theorem B Abbott Sec. 6.5

Power series converge uniformly inside the radius

If $0 < r < R$ , then $\sum c_k x^k$ converges uniformly on $[-r, r]$ , and its sum is a continuous function on $(-R, R)$ .

In one lineIf $0 < r < R$ , then $\sum c_k x^k$ converges uniformly on $[-r, r]$ , so its sum is continuous on $(-R, R)$ .

Why it is trueOn [-r, r] the terms are bounded by |c_k| r^k, and the series of these bounds converges because r is inside the radius; then the M-test applies.

ProofAbbott 6.5.3

The idea: M-test with M_k = |c_k| r^k, which sums because the series converges absolutely at r < R.

\sum |c_k| r^k

converges.

r < R

, so the series converges absolutely at

x = r

(definition of the radius).

The M-test applies with

M_k = |c_k| r^k

, so the convergence is uniform on

[-r, r]

For

|x| \le r

|c_k x^k| \le |c_k| r^k = M_k

So the sum is continuous on

(-R, R)

Partial sums are polynomials, which are continuous; uniform limits of continuous functions are continuous. Every point of

(-R, R)

lies in some

[-r, r]

\blacksquare

Integrating and differentiating term by term

Theorem B Abbott Thm 6.3.1

Integrating a uniform limit

If each $f_n$ is continuous on $[a, b]$ and $f_n \to f$ uniformly, then $f$ is continuous (hence integrable by 1.5) and $\int_a^b f_n \to \int_a^b f$ .

In one lineIf continuous $f_n \to f$ uniformly on $[a, b]$ , then $\int_a^b f_n \to \int_a^b f$ .

Why it is trueThe area between f_n and f is at most the height of the uniform tube times the width b − a.

ProofAbbott 6.3.1

The idea: |∫f_n − ∫f| ≤ ∫|f_n − f| ≤ (b − a)·sup|f_n − f|, which tends to 0 by uniform convergence.

f

is continuous on

[a, b]

, so integrable.

Uniform limit of continuous functions is continuous, and continuous functions are integrable (1.5).

\left|\int_a^b f_n - \int_a^b f\right| = \left|\int_a^b (f_n - f)\right|

Linearity of the integral (1.5).

For

n > N

\left|\int_a^b (f_n - f)\right| \le \dfrac{\varepsilon}{b - a}(b - a) = \varepsilon

\blacksquare

Given

\varepsilon > 0

choose

N

with

|f_n - f| < \frac{\varepsilon}{b - a}

everywhere for

n > N

(uniform convergence). Then

-\frac\varepsilon{b-a} \le f_n - f \le \frac\varepsilon{b-a}

, and monotonicity of the integral (1.5).

Theorem B Abbott Thm 6.3.3

Differentiating a uniform limit of derivatives

Let $f_n$ be differentiable on $[a, b]$ with continuous $f_n'$ . If $f_n' \to g$ uniformly on $[a, b]$ and $f_n(a) \to c$ , then $f_n \to f$ pointwise, where $f$ is differentiable and $f' = g$ .

In one lineIf $f_n$ have continuous derivatives, $f_n' \to g$ uniformly and $f_n(a)$ converges, then $f_n \to f$ with $f' = g$ .

Why it is trueWrite f_n as its starting value plus the integral of its derivative; integrate the uniform limit term by term, then differentiate with the fundamental theorem.

ProofAbbott 6.3.3 (proof via the integral)

The idea: By the fundamental theorem part 2, f_n(x) = f_n(a) + ∫_a^x f_n'. Pass to the limit with the integration theorem to get f(x) = c + ∫_a^x g, then use part 1.

f_n(x) = f_n(a) + \int_a^x f_n'(t)\,dt

for every

x \in [a, b]

Fundamental theorem of calculus, part 2, with

f_n'

continuous (so integrable).

\int_a^x f_n' \to \int_a^x g

g

is a uniform limit of continuous functions, hence continuous; the integration theorem applied on

[a, x]

f_n(x) \to c + \int_a^x g =: f(x)

Limit laws of 1.2.

f'(x) = g(x)

Fundamental theorem of calculus, part 1, since

g

is continuous.

\blacksquare

Why· The reason for the hypothesis on derivatives

Uniform convergence of $f_n$ alone does not allow differentiating the limit. Take $f_n(x) = \frac{\sin nx}{n}$ : it tends to $0$ uniformly (since $|f_n| \le \frac1n$ ), so the limit has derivative $0$ , but $f_n'(x) = \cos nx$ has no limit at all. The hypothesis must be placed on the derivatives $f_n'$ , as in the theorem.

Beyond the course: for understanding· Termwise differentiation of power series (label B)

Apply the two theorems to the partial sums of a power series whose coefficient ratios tend to $L$ . The derived series $\sum k c_k x^{k-1}$ has coefficient ratio $\frac{(k+1)|c_{k+1}|}{k|c_k|} \to L$ , so it has the same radius $R$ . On each $[-r, r]$ with $r < R$ both $\sum c_k x^k$ and $\sum k c_k x^{k-1}$ converge uniformly, so inside $(-R, R)$ : the sum $f(x) = \sum c_k x^k$ is differentiable with $f'(x) = \sum_{k \ge 1} k c_k x^{k-1}$ , and $\int_0^x f = \sum \frac{c_k}{k+1}x^{k+1}$ . (For general coefficients the radius is given by the Cauchy–Hadamard formula $\frac1R = \limsup |c_k|^{1/k}$ , and the same conclusions hold.) So inside the radius a power series behaves like a polynomial that never ends.

Worked example Abbott 6.5 style

Find

\sum_{k=1}^\infty \frac{k}{2^k}

by differentiating the geometric series.

Solution.

Start with the geometric series. $\sum_{k=0}^\infty x^k = \frac1{1-x}$ for $|x| < 1$ . Radius $R = 1$ .
Differentiate both sides, term by term. $\sum_{k=1}^\infty k x^{k-1} = \frac1{(1-x)^2}$ . Allowed inside the radius of convergence, by the termwise differentiation theorem.
Multiply by $x$ . $\sum_{k=1}^\infty k x^k = \frac{x}{(1-x)^2}$ . A constant (or a factor $x$ ) multiplies each term.
Put $x = \frac12$ . $\sum \frac{k}{2^k} = \dfrac{1/2}{(1/2)^2} = 2$ . $\frac12$ is inside $(-1, 1)$ .

Answer. $2$ . $\blacksquare$ The first partial sums are $0.5, 1.0, 1.375, 1.625, \dots$ , creeping to $2$ .

Your answerUse the same method to find

\displaystyle\sum_{k=1}^\infty \frac{k}{3^k}

$e^x$ , $\sin$ , $\cos$ , and Euler’s formula

Define, for every real $x$ ,

E(x) = \sum_{k=0}^\infty \frac{x^k}{k!}, \qquad S(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}, \qquad C(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}.

The ratio test shows each converges absolutely for every real $x$ (for $E$ the ratio is $\frac{|x|}{k+1} \to 0$ ; for $S$ and $C$ it is $\frac{x^2}{(2k+2)(2k+3)}$ or $\frac{x^2}{(2k+1)(2k+2)}$ , also tending to $0$ ). So $R = \infty$ .

Theorem B Abbott Sec. 6.6; Rudin ch. 8

Properties of the exponential series

For all real $x, y$ : $E'(x) = E(x)$ , $S'(x) = C(x)$ , $C'(x) = -S(x)$ , $E(0) = 1$ , and $E(x + y) = E(x)E(y)$ .

In one line $E' = E$ , $S' = C$ , $C' = -S$ , and $E(x + y) = E(x)\,E(y)$ , so $E(x) = e^x$ with $e = E(1) = 2.71828\ldots$

Why it is trueTermwise differentiation shows the series solve the differential equations that define exp, sin and cos, and the product rule for series gives the law of exponents.

ProofAbbott Sec. 6.6; Rudin ch. 8

The idea: Differentiate term by term: the exponent k comes down and cancels one factor in k!. The law of exponents needs the product of two absolutely convergent series (Cauchy product), proved after.

E'(x) = \sum_{k \ge 1} \dfrac{x^{k-1}}{(k-1)!} = E(x)

Termwise differentiation inside

R = \infty

\frac{d}{dx}\frac{x^k}{k!} = \frac{k x^{k-1}}{k!} = \frac{x^{k-1}}{(k-1)!}

S' = C

and

C' = -S

The same, term by term: the derivative of

\frac{(-1)^k x^{2k+1}}{(2k+1)!}

\frac{(-1)^k x^{2k}}{(2k)!}

; the derivative of

\frac{(-1)^k x^{2k}}{(2k)!}

\frac{(-1)^k x^{2k-1}}{(2k-1)!}

, which is the term of

-S

with

k-1

in place of

k

. These series have zero coefficients (only even or only odd powers), so the ratio formula does not apply as written. Instead use the M-test directly: on

[-r, r]

every term of

S

C

and of their term-by-term derivatives is at most

\frac{r^n}{n!}

for the matching

n

, and

\sum \frac{r^n}{n!}

converges by the ratio test. So all of them converge uniformly on

[-r, r]

, and the two differentiation theorems apply.

E(0) = 1

S(0) = 0

C(0) = 1

Put

x = 0

in each series: only the first term survives.

E(x)E(y) = \sum_n \sum_{j + k = n} \dfrac{x^j y^k}{j!\,k!} = \sum_n \dfrac{(x+y)^n}{n!} = E(x+y)

, by the binomial theorem.

\blacksquare

Cauchy product (next theorem), with

a_j = \frac{x^j}{j!}

and

b_k = \frac{y^k}{k!}

, both absolutely convergent. The sum over

j + k = n

\frac1{n!}\sum \binom nj x^j y^{n-j}

Theorem B Rudin Thm 3.50 (Cauchy product)

Product of absolutely convergent series

If $\sum_{j \ge 0} a_j = A$ and $\sum_{k \ge 0} b_k = B$ converge absolutely and $c_n = \sum_{j=0}^n a_j b_{n-j}$ , then $\sum c_n = AB$ .

In one lineIf $\sum a_j$ and $\sum b_k$ converge absolutely, then $\sum_n c_n$ with $c_n = \sum_{j+k=n} a_j b_k$ converges to $\left(\sum a_j\right)\left(\sum b_k\right)$ .

Why it is trueThe partial products and the partial sums of c differ only by terms a_j b_k where j or k is large, and those are tiny in absolute value.

ProofRudin 3.50

The idea: A_nB_n is the sum of a_j b_k over the square j,k ≤ n, and C_n the sum over the triangle j + k ≤ n. The difference is the sum over the corner j + k > n with j, k ≤ n, where j or k exceeds n/2, so it is at most the product of a tail with the total.

A_nB_n - C_n = \sum_{j,k \le n,\ j + k > n} a_j b_k

Expand the product of partial sums: the square contains the triangle.

Each such pair has

j > \frac n2

k > \frac n2

j + k > n

then

j > \frac n2

k > \frac n2

(otherwise

j + k \le n

|A_nB_n - C_n| \le \left(\sum_{j > n/2}|a_j|\right)\beta + \alpha\left(\sum_{k > n/2}|b_k|\right)

Triangle inequality, then bound the double sum by sums over a row times a column. Let

\alpha = \sum|a_j|

\beta = \sum|b_k|

The right side tends to

0

, so

C_n \to AB

\blacksquare

Tails of convergent series tend to

0

. Also

A_nB_n \to AB

by the product law of 1.2.

Since $E(x)E(-x) = E(0) = 1$ , $E$ is never $0$ , and $E(x) = E(\frac x2)^2 > 0$ . Since $E' = E > 0$ , $E$ is increasing. So $E$ is the familiar $e^x$ : the law of exponents gives $E(n) = E(1)^n$ for whole $n$ and $E\left(\frac pq\right)^q = E(p) = E(1)^p$ for fractions, so $E(x) = e^x$ at every rational $x$ with $e = E(1)$ , and a differentiable (hence continuous) function is fixed by its values on the rationals. (Also, $E$ is the unique function with $E' = E$ and $E(0) = 1$ ; a stretch problem.) The numbers: $\sum_{k=0}^{10}\frac1{k!} = 2.7182818011$ against $e = 2.7182818285$ , an error of $2.7 \times 10^{-8}$ .

Euler’s formula

Series with complex terms behave just as real ones do, because the complex numbers are also complete (a Cauchy sequence of complex numbers converges, by applying the real result to real and imaginary parts, since $|\operatorname{Re} z|, |\operatorname{Im} z| \le |z|$ ), and the triangle inequality $|z + w| \le |z| + |w|$ holds. So everything above (absolute convergence, ratio test, Cauchy product) works verbatim, with $E(z) = \sum \frac{z^k}{k!}$ for complex $z$ , and $E(z + w) = E(z)E(w)$ holds by the same proof.

The series $C$ and $S$ were defined by formulas, but $\cos$ and $\sin$ in school come from the unit circle. The next lemma shows they are the same functions. It uses one outside fact, which is taught in school calculus and not proved in this book: the circle functions satisfy $\sin' = \cos$ , $\cos' = -\sin$ , $\sin 0 = 0$ , $\cos 0 = 1$ .

Lemma B new

The series C and S are cosine and sine

Assume the school facts above. Then $C = \cos$ and $S = \sin$ on $\mathbb R$ .

In one lineFor every real $x$ , $C(x) = \cos x$ and $S(x) = \sin x$ , where $\cos$ and $\sin$ are the unit-circle functions.

Why it is trueTwo pairs of functions that start at the same values and obey the same differential equations cannot drift apart, because the quantity u squared plus v squared for their differences has derivative zero.

Proofnew

The idea: Look at the differences u = C − cos and v = S − sin. They start at 0, and u' = −v, v' = u. Then u² + v² has derivative 0, so it stays 0.

Let

u = C - \cos

and

v = S - \sin

Define the differences. Both

C, S

(from the previous theorem) and

\cos, \sin

(school fact) are differentiable.

Then

u' = -v

and

v' = u

C' = -S

and

\cos' = -\sin

, so

u' = -S + \sin = -v

. Also

S' = C

and

\sin' = \cos

, so

v' = C - \cos = u

u(0) = 0

and

v(0) = 0

Put

x = 0

C(0) = 1 = \cos 0

and

S(0) = 0 = \sin 0

Let

w = u^2 + v^2

. Then

w' = 2uu' + 2vv' = -2uv + 2vu = 0

Product and chain rules:

(u^2)' = 2uu'

and

(v^2)' = 2vv'

. Then substitute step 2.This is the idea: find a quantity that cannot change. It is like energy in physics.

w

is constant, and

w(0) = 0

. Hence

u^2 + v^2 = 0

for every

x

A function with zero derivative on an interval is constant (1.4, a corollary of the mean value theorem).

u = 0

and

v = 0

, that is

C = \cos

and

S = \sin

\blacksquare

A sum of two squares of real numbers is

0

only if each is

0

Theorem B Euler 1748; Rudin ch. 8

Euler's formula via power series

For every real $\theta$ : $E(i\theta) = C(\theta) + i\,S(\theta)$ . Since $C$ and $S$ are the cosine and sine series, $e^{i\theta} = \cos\theta + i\sin\theta$ . In particular $e^{i\pi} = -1$ .

In one line $e^{i\theta} = \cos\theta + i\sin\theta$ for every real $\theta$ , where $e^{z} = \sum z^k/k!$ .

Why it is trueThe powers of i cycle through 1, i, -1, -i, which sorts the exponential series into the series of cosine (even powers) and sine (odd powers).

ProofEuler 1748; Rudin ch. 8

The idea: Put z = iθ in the series and sort the terms by the remainder of the exponent mod 4. The partial sum with an odd number of terms splits exactly into a cosine partial sum and i times a sine partial sum.

The term

\frac{(i\theta)^n}{n!}

equals

\frac{(-1)^k\theta^{2k}}{(2k)!}

for

n = 2k

and

i\,\frac{(-1)^k\theta^{2k+1}}{(2k+1)!}

for

n = 2k + 1

i^2 = -1

i^3 = -i

i^4 = 1

, then the cycle repeats:

i^{2k} = (-1)^k

and

i^{2k+1} = (-1)^k i

The partial sum is

\sum_{n \le 2m+1}\dfrac{(i\theta)^n}{n!} = \sum_{k=0}^{m}\dfrac{(-1)^k\theta^{2k}}{(2k)!} + i\sum_{k=0}^{m}\dfrac{(-1)^k\theta^{2k+1}}{(2k+1)!}

Finite sums can be regrouped freely: add the first

2m + 2

terms (

n = 0, \dots, 2m+1

), even ones together and odd ones together.

Letting

m \to \infty

E(i\theta) = C(\theta) + i\,S(\theta)

The left side tends to

E(i\theta)

, a subsequence of the convergent partial sums (1.2); the sums on the right tend to

C(\theta)

and

S(\theta)

by their own convergence; limit laws of 1.2. No rearrangement of an infinite series was needed.

e^{i\theta} = \cos\theta + i\sin\theta

. With

\theta = \pi

e^{i\pi} = -1 + 0\cdot i = -1

\blacksquare

The lemma just proved:

C = \cos

and

S = \sin

The identification of $C$ and $S$ with the school $\cos$ and $\sin$ is the lemma just before the theorem; its only outside input is the school fact that the geometric $\sin$ and $\cos$ have derivatives $\cos$ and $-\sin$ . This is the proof promised in 0.6, and the multiplication rule there, $re^{i\alpha}\cdot se^{i\beta} = rs\,e^{i(\alpha+\beta)}$ , is the law of exponents $E(z + w) = E(z)E(w)$ proved above.

Worked example

Use the series to compute

e^{i\pi/2}

, and to compute

\sin(0.5)

with an error bound.

Solution.

Euler’s formula. $e^{i\pi/2} = \cos\frac\pi2 + i\sin\frac\pi2 = 0 + i = i$ . Substitute $\theta = \frac\pi2$ .
Check with the law of exponents. $\left(e^{i\pi/2}\right)^2 = e^{i\pi} = -1 = i^2$ . $E(z + w) = E(z)E(w)$ with $z = w = i\frac\pi2$ .
Series for $\sin 0.5$ . $S(0.5) \approx 0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} = 0.5 - 0.0208333 + 0.0002604 = 0.4794271$ . Three terms of the sine series.
Error bound. The series alternates with decreasing terms, so the error is at most the next term, $\frac{0.5^7}{5040} = 1.55 \times 10^{-6}$ . The true value is $0.4794255$ , an error of $1.54 \times 10^{-6}$ . Alternating series test error bound.

Answer. $e^{i\pi/2} = i$ , and $\sin 0.5 = 0.47943$ to five places from three terms. $\blacksquare$

Where it is used

Used inside maths· Inside maths: solving equations, analytic functions, Fourier series

To solve $y' = y$ , try $y = \sum c_k x^k$ . Differentiating term by term (allowed inside the radius) gives $\sum k c_k x^{k-1} = \sum c_k x^k$ , so $(k+1)c_{k+1} = c_k$ and $c_k = \frac{c_0}{k!}$ : the series for $e^x$ appears without being guessed. The same method solves many differential equations of MATH 376. A function that equals its power series near every point is called analytic; complex analysis shows that complex-differentiable functions are always analytic, and the radius of convergence is the distance to the nearest singularity. The uniform limit theorem is the reason why the Fourier series of a function with a jump cannot converge uniformly (Abel’s example), which leads to the notion of convergence used in the next application.

Used in the real world· Real world: calculators, signal processing, and the number e

Calculators and floating point. A calculator reduces the angle to $[-\frac\pi4, \frac\pi4]$ and then sums a Taylor polynomial. For $\sin(\frac\pi4)$ , three terms plus one more give $0.70710647$ against the true $0.70710678$ ; the error $3.1 \times 10^{-7}$ is just below the first omitted term $\frac{x^9}{9!} = 3.13 \times 10^{-7}$ , as the alternating series test predicts. A double-precision number near $0.7$ has spacing about $1.1 \times 10^{-16}$ between neighbours, and the first omitted term $\frac{x^{17}}{17!}$ at $x = \frac\pi4$ is $4.6 \times 10^{-17}$ , below that. So eight terms of the sine series (through $x^{15}$ ) suffice for full double precision on that interval (real libraries use polynomials tuned slightly differently, but the series explains the degree).

Signal processing. The sawtooth $f(x) = x$ on $(-\pi, \pi)$ has the Fourier series $2\sum_{k \ge 1}\frac{(-1)^{k+1}}{k}\sin kx$ . It cannot converge uniformly, because the sum would be continuous but $f$ jumps between $\pi$ and $-\pi$ at the ends of the period. It converges in energy: the squared error after $n$ terms is $\int |f - S_n|^2 = 4\pi\sum_{k > n}\frac1{k^2}$ , which is $1.196$ for $n = 10$ and $0.125$ for $n = 100$ , out of a total energy $\int f^2 = \frac{2\pi^3}{3} = 20.67$ . So 10 harmonics already capture $94\%$ of the energy and 100 capture $99.4\%$ . The mean-square sense of convergence is made precise in 2.9, inner products of functions.

Compound interest and $e$ . One unit of money at $100\%$ annual interest compounded $n$ times a year grows to $\left(1 + \frac1n\right)^n$ : $2$ for yearly, $2.6130$ monthly, $2.7146$ daily, and in the limit of continuous compounding $e = \sum\frac1{k!} = 2.7183$ . The series converges so fast (the terms up to $\frac1{10!}$ give $2.7182818$ , eight correct digits) that it is how $e$ is computed in practice.

Traps

Watch out· Terms tending to zero does not make a series converge

$\frac1k \to 0$ and $\sum\frac1k$ diverges. The nth-term test only proves divergence: if $a_k \not\to 0$ , the series diverges; if $a_k \to 0$ , no conclusion.

Watch out· Do not rearrange conditionally convergent series

The alternating harmonic series sums to $\ln 2$ in its usual order and to $\frac12\ln 2$ in the order one positive, two negative. Reordering, regrouping infinite sums or swapping the order of a double sum is safe only for absolutely convergent series.

Watch out· Ratio of 1 means the ratio test is silent

$\sum\frac1k$ and $\sum\frac1{k^2}$ both have ratio tending to $1$ , and one diverges while the other converges. Use another test.

Watch out· Pointwise limits can destroy continuity, integrals and derivatives

For $f_n(x) = x^n$ , the limit of the continuous functions is discontinuous. For tent functions $f_n$ of height $n$ on $[0, \frac1n]$ (zero elsewhere), $f_n \to 0$ pointwise but $\int_0^1 f_n = \frac12$ for every $n$ , so $\lim\int f_n \ne \int\lim f_n$ . For $f_n = \frac{\sin nx}{n}$ , $f_n \to 0$ but $f_n'$ does not converge. All three are repaired by uniform convergence (of the function, or of the derivatives).

Watch out· At the endpoints of the interval of convergence, check separately

The radius gives convergence for $|x| < R$ and divergence for $|x| > R$ , nothing at $|x| = R$ . The series $\sum\frac{x^k}{k}$ has $R = 1$ , diverges at $x = 1$ and converges at $x = -1$ .

Problems

check

Does $\sum_{k=1}^\infty \frac{k}{k+1}$ converge?

Solution

No. The terms $\frac{k}{k+1} \to 1 \ne 0$ , so by the nth-term test the series diverges.

check

Is $\sum\frac{(-1)^k}{k}$ absolutely convergent, conditionally convergent, or divergent?

Solution

Conditionally convergent: the alternating series test gives convergence, but $\sum\left|\frac{(-1)^k}{k}\right| = \sum\frac1k$ diverges.

check

Let $f_n(x) = \frac xn$ . Find the pointwise limit on $\mathbb R$ . Is the convergence uniform on $\mathbb R$ ? On $[-5, 5]$ ?

Solution

The pointwise limit is $f = 0$ , since $\frac xn \to 0$ for each fixed $x$ . On $\mathbb R$ it is not uniform: $\sup_x\left|\frac xn\right| = \infty$ (take $x = n$ , giving $1$ for every $n$ ). On $[-5, 5]$ it is uniform: $\sup\left|\frac xn\right| = \frac5n \to 0$ .

use

Use the ratio test on $\sum\frac{k!}{k^k}$ .

Solution

$\dfrac{a_{k+1}}{a_k} = \dfrac{(k+1)!}{(k+1)^{k+1}}\cdot\dfrac{k^k}{k!} = \dfrac{k^k}{(k+1)^k} = \left(1 + \dfrac1k\right)^{-k} \to \dfrac1e = 0.368 < 1$ (using $(1 + \frac1k)^k \to e$ ). So the series converges. (Numerically the ratio at $k = 100$ is $0.3697$ .)

use

Find the radius of convergence and the exact set of convergence of $\sum\frac{x^k}{k^2}$ .

Solution

Coefficient ratio $\frac{k^2}{(k+1)^2} \to 1$ , so $R = 1$ . At $x = \pm1$ the series is $\sum\frac{(\pm1)^k}{k^2}$ , which converges absolutely (comparison with $\sum\frac1{k^2}$ ). So the set of convergence is $[-1, 1]$ .

use

Find $\sum_{k=1}^\infty\frac1{k(k+1)}$ and $\sum_{k=0}^\infty\left(-\frac12\right)^k$ .

Solution

First: $\frac1{k(k+1)} = \frac1k - \frac1{k+1}$ , so $S_n = 1 - \frac1{n+1} \to 1$ . Second: geometric series with $r = -\frac12$ : $\frac1{1 - (-1/2)} = \frac23$ .

use

Show that $\sum_{k=1}^\infty\frac{\cos kx}{k^2}$ converges uniformly on $\mathbb R$ to a continuous function.

Solution

$\left|\frac{\cos kx}{k^2}\right| \le \frac1{k^2}$ , and $\sum\frac1{k^2}$ converges, so the M-test gives uniform convergence. The partial sums are continuous, so the limit is continuous.

proveAbbott Sec. 2.7 (p-series)

Prove that $\sum\frac1{k^p}$ converges if $p > 1$ and diverges if $p \le 1$ .

Solution

If $p \le 1$ : $\frac1{k^p} \ge \frac1k$ for $k \ge 1$ , so the series diverges by comparison with the harmonic series. If $p > 1$ : group the terms in blocks $2^j \le k < 2^{j+1}$ . The block has $2^j$ terms, each at most $\frac1{2^{jp}}$ , so it adds to at most $2^j\cdot 2^{-jp} = \left(2^{1-p}\right)^j$ . Since $p > 1$ , $2^{1-p} < 1$ , and these block bounds form a convergent geometric series. Every partial sum is at most the sum of finitely many block bounds, which is bounded; the terms are positive, so the partial sums increase and are bounded, and the series converges by monotone convergence.

prove

Prove that for an alternating series with $b_k$ decreasing to $0$ , the sum $S$ lies between any two consecutive partial sums, and hence $|S - S_n| \le b_{n+1}$ .

Solution

Even partial sums increase to $S$ and odd ones decrease to $S$ (proof of the alternating series test). So for each $n$ , $S$ is between $S_n$ and $S_{n+1}$ (one is even-indexed, the other odd-indexed). Then $|S - S_n| \le |S_{n+1} - S_n| = b_{n+1}$ .

proveAbbott Sec. 6.2

Prove: $f_n \to f$ uniformly on $D$ if and only if $\sup_{x\in D}|f_n(x) - f(x)| \to 0$ (when each supremum is finite).

Solution

Let $s_n = \sup_D|f_n - f|$ . ( $\Rightarrow$ ) Given $\varepsilon$ , uniform convergence gives $N$ with $|f_n(x) - f(x)| < \frac\varepsilon2$ for all $x$ and all $n > N$ . So $s_n \le \frac\varepsilon2 < \varepsilon$ , that is $s_n \to 0$ . ( $\Leftarrow$ ) Given $\varepsilon$ , there is $N$ with $s_n < \varepsilon$ for $n > N$ ; then $|f_n(x) - f(x)| \le s_n < \varepsilon$ for every $x$ . (The $\frac\varepsilon2$ in the first part is needed because a supremum of values below $\varepsilon$ may equal $\varepsilon$ .)

proveAbbott Sec. 6.2 exercises

Prove: if $f_n \to f$ uniformly on $D$ and every $f_n$ is uniformly continuous on $D$ , then $f$ is uniformly continuous on $D$ .

Solution

prove

Prove that $\sum x^k$ converges uniformly on $[-r, r]$ for each $r < 1$ but not on $(-1, 1)$ .

Solution

On $[-r, r]$ : $|x^k| \le r^k$ and $\sum r^k$ converges, so the M-test gives uniform convergence. On $(-1, 1)$ : the limit is $\frac1{1-x}$ and the partial sum is $\frac{1 - x^{n+1}}{1-x}$ , so the error is $\frac{x^{n+1}}{1-x}$ . For fixed $n$ the error tends to $\infty$ as $x \to 1^-$ (the numerator tends to $1$ and the denominator to $0$ ), so $\sup_{(-1,1)}\left|\frac{x^{n+1}}{1-x}\right| = \infty$ and does not tend to $0$ . Not uniform. (Another proof: the partial sums are continuous and bounded on $(-1,1)$ but the limit $\frac1{1-x}$ is unbounded there, and a uniform limit of bounded functions is bounded.)

prove

Prove that the limit function in $f(x) = \sum_{k=0}^\infty\frac{x^k}{k!}$ satisfies $f(x) > 0$ for all real $x$ and $f$ is strictly increasing.

Solution

By the law of exponents, $f(x)f(-x) = f(0) = 1$ , so $f(x) \ne 0$ for every $x$ . Also $f(x) = f(\frac x2)^2 \ge 0$ and it is not $0$ , hence $f(x) > 0$ . Differentiating term by term, $f' = f > 0$ , so $f$ is strictly increasing (by the mean value theorem, a function with positive derivative is strictly increasing).

stretchRudin Exercise 3.50 style (Cauchy product of conditionally convergent series)

Let $a_k = \frac{(-1)^k}{\sqrt k}$ for $k \ge 1$ . Show that $\sum a_k$ converges, but its Cauchy product with itself, $c_n = \sum_{j=1}^{n-1}a_j a_{n-j}$ , diverges. So absolute convergence in the product theorem cannot simply be dropped.

Solution

Convergence: the alternating series test with $b_k = \frac1{\sqrt k}$ decreasing to $0$ . Product: $c_n = (-1)^n\sum_{j=1}^{n-1}\frac1{\sqrt{j(n-j)}}$ . For $1 \le j \le n-1$ , $j(n-j) \le \frac{n^2}{4}$ (AM-GM), so each term is at least $\frac2n$ , and there are $n - 1$ terms. So $|c_n| \ge \frac{2(n-1)}{n} \ge 1$ for $n \ge 2$ . Thus $c_n \not\to 0$ , and $\sum c_n$ diverges by the nth-term test.

stretchAbbott Sec. 6.6

Let $y$ be differentiable on $\mathbb R$ with $y' = y$ and $y(0) = 1$ . Prove $y(x) = e^x$ (where $e^x = E(x)$ is the power series).

Solution

Let $g(x) = y(x)E(-x)$ . Then $g'(x) = y'(x)E(-x) - y(x)E'(-x) = y(x)E(-x) - y(x)E(-x) = 0$ (product and chain rule, $E' = E$ ). So $g$ is constant (mean value theorem), and $g(0) = y(0)E(0) = 1$ . So $y(x)E(-x) = 1$ . Multiply by $E(x)$ and use $E(x)E(-x) = 1$ : $y(x) = E(x)$ .

Summary

Series are limits of partial sums. Geometric series: $\sum r^k = \frac1{1-r}$ for $|r| < 1$ . Terms tending to $0$ is necessary, not sufficient (harmonic series, Oresme).
Tests: comparison (increasing partial sums, bounded by the bigger series), ratio (compare with a geometric series; silent at $r = 1$ ), alternating (even sums up, odd sums down, error at most the next term).
Absolute convergence implies convergence (Cauchy criterion plus the triangle inequality) and allows any rearrangement. Conditionally convergent series can be rearranged to any sum (Riemann).
Pointwise versus uniform: uniform means one $N$ for all $x$ ( $\sup|f_n - f| \to 0$ ). A uniform limit of continuous functions is continuous (the $\frac\varepsilon3$ proof); $x^n$ on $[0, 1]$ shows pointwise is not enough.
Weierstrass M-test: bound $|f_k| \le M_k$ with $\sum M_k < \infty$ and the series converges uniformly.
Power series have a radius $R = \frac1L$ (ratio formula), converge uniformly inside, and can be integrated and differentiated term by term there (via uniform convergence and the fundamental theorem).
$e^x$ , $\sin$ , $\cos$ are power series with $R = \infty$ ; $E' = E$ and $E(z + w) = E(z)E(w)$ (Cauchy product), and Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ follows by sorting the series by powers of $i$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

1.6 Series, power series, and uniform convergence

The picture: two ways to converge

The idea, in words

Series: definitions and first tests

Series

Partial sum

Geometric series

The nth-term test

The harmonic series diverges

Comparison test

Absolute convergence

Absolute convergence

Cauchy criterion for series

Absolute convergence implies convergence

Ratio test

Alternating series test

Rearrangements

Rearranging an absolutely convergent series

Sequences of functions: pointwise versus uniform

Pointwise convergence

Uniform convergence

Uniform limit of continuous functions is continuous

Weierstrass M-test

Power series

Power series

Convergence of power series (Abel's lemma)

Radius of convergence

Ratio formula for the radius

Power series converge uniformly inside the radius

Integrating and differentiating term by term

Integrating a uniform limit

Differentiating a uniform limit of derivatives

exe^xex, sin⁡\sinsin, cos⁡\coscos, and Euler’s formula

Properties of the exponential series

Product of absolutely convergent series

Euler’s formula

The series C and S are cosine and sine

Euler's formula via power series

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere

$e^x$ , $\sin$ , $\cos$ , and Euler’s formula