$$$$

17  Power Series

Chapter highlights: we prove that a power series is continuous on its entire domain: this is a combination of two theorems, that (1) its continuous on the interior and (2) that its continuous at boundary points, when defined there. Proving the continuity of series provides an opportunity to use the material we learned in Series: Advanced Techniques. In particular, we will use Dominated Convergence to prove the continuity of series within the interval of convergence, and Summation By Parts to prove continuity at any boundary points.

17.1 Continuity In the Interior

We have proven previously that a power series $\sum _{n\ge 0}{a}_n{x}^n$ either converges only at $\{0\}$, converges on the entire real line, or has a finite radius of convergence, where it converges on an interval of the form $(-r,r)$, $[r,r]$, $(-r,r]$ or $[-r,r)$. Plotting the partial sums of such a series shows that outside the radius things quickly blow up to infinity, but within the radius of convergence the result appears to be continuous. We confirm this below.

Theorem 17.1 (Continuity within Radius of Convergence) Let $f(x)=\sum _k{a}_k{x}^k$ be a power series with radius of convergence $r$. Then if $|x|<r$, $f$ is continuous at $x$.

Proof. Take $x>0$ (leaving the trivial modifications for $x<0$ as an exercise), and let $x_n$ be an arbitrary sequence in $(-r,r)$ converging to $x$. We aim to show that $f(x_n)\to f(x)$.

As $x<r$ choose some $y$ with $x<y<r$ (perhaps, $y=(x+r)/2$). Since $x_n\to x$ there is some $N$ past which $x_n$ is always less than $y$ (take $\epsilon =y-x$ and apply the definition of $x_n\to x$). As truncating the terms of the sequence before this does not change its limit, we may without loss of generality assume that $x_n<y$ for all $n$. Thus, we may define $M_k=a_k{y}^k$, and we are in a situation to verify the hypotheses of Dominated Convergence:

• Since $x_n\to x$, we have $a_k{x}_n^k\to a_k{x}^k$ by the limit theorems.
• For each $n$, $f(x_n)=\sum _k{a}_k{x}_n^k$ is convergent as $x_n$ is within the radius of convergence.
• $M_k=a_k{y}^k$ bounds $a_k{x}_n^k$ for all $n$, as $0<x_n<y$.
• $\sum _k{M}_k$ converges as this is just $f(y)$ and $y$ is within the radius of convergence.

Applying the theorem, we see $$\underset{n}{lim}f(x_n)=\underset{n}{lim}\sum _k{a}_k{x}_n^k=\sum _{k}lim{a}_k{x}_n^k=\sum _k{a}_k{x}^k=f(x)$$

Thus for arbitrary $x_n\to x$ we have $f(x_n)\to f(x)$, so $f$ is continuous at $x$.

Remark 17.1. If the power series converges on all of $\mathbb{R}$, the same proof above holds, taking a sequence $x_n\to x$, but its a little easier: we don’t have to be careful choosing our upper bound $y$. Any upper bound for the convergent sequence $\{x_n\}$ will do.

We pause to remark this result is something rather special to power series, and is not true in general: its quite possible to write down a sequence of continuous functions which converges to a discontinuous function. So the fact a sequence of (continuous) partial sums of a power series converges to a continuous limit is indeed a big deal! This is one of many things that makes power series particularly nice.

Exercise 17.1 Let $f(x)=\frac{x}{\sqrt{x^2+1}}$, and define the sequence of functions $f_n(x)=f(nx)$. As $n\to \mathrm{\infty }$ prove that

• If $x>0$ then $f_n(x)\to 1$
• If $x=0$ then $f_n(x)\to 0$
• If $x<0$ then $f_n(x)\to -1$

Thus, while $f_n$ is continuous for each $n$ (its a composition of continuous functions), the limit is discontinuous.

There is a lot of theoretical work in real analysis to determine more general conditions under which a sequence of continuous functions converges to a continuous limit. In this semester long course we won’t have need for such results beyond the power series case above, but in the eventual extension of this book, we will develop the notion of uniform convergence for this purpose.

17.2 $⧫$ Continuity at the Boundary

While we now completely understand a power series on the interior of its radius of convergence, there’s a little more work to do to complete the picture.

Theorem 17.2 (Continuity at the Boundary: Abel’s Theorem) Let $f(x)=\sum _{k\ge 0}{a}_k{x}^k$ be a power series with radius of convergence $r$, which converges at an endpoint $\pm r$. Then $f$ is continuous there.

The full proof of this theorem is rather more technical than the previous result, and before proving it through a sequence of steps, we pause to appreciate why. First, note that some cases of this theorem really are easy: for instance, if the endpoint converges absolutely, you can carry out the exact same proof using Dominated Convergence as above.

Exercise 17.2 Let $\sum _{k\ge 0}{a}_k{x}^k$ be a power series with radius of convergence $1$, and suppose $\sum _{k\ge 0}{a}_k$ converges absolutely. Then $$\underset{x\to 1^{-}}{lim}\sum _{k\ge 0}{a}_k{x}^k=\sum _{k\ge 0}{a}_k$$

The difficulty is then is what happens when $\sum _{k\ge 0}{a}_k$ converges, but does not do so absolutely. This is a real case, that actually shows up in important situations rather often: for example the power series

$$\sum _{n\ge 0}\frac{(-1{)}^n}{n+1}{x}^{n+1}\sum _{n\ge 0}\frac{(-1{)}^n}{2n+1}{x}^{2n+1}$$

Both converge at $x=1$, but their absolute values diverge. However, the continuity of these power series at their endpoints will be absolutely essential to us later on in the course, when deriving the amazing identities

$$\log (2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

We also pause to quickly dash any hopes there might be a general sort of proof: (perhaps one hopes that if a sequence of functions converges on $(-1,1)$ to a continuous limit, and also converges at a boundary point then it is automatically continuous at the boundary).

Example 17.1 Consider the sequence $f_n(x)=x^n$. As $n\to \mathrm{\infty }$ this converges on the interval $(-1,1]$ and diverges everywhere else. Furthermore for $|x|<1$ it converges to the zero function which is constant - hence continuous. However at $x=1$ the sequence $f_n(1)=1^n=1$ is constantly equal to 1, so the limit is also $1$. Thus the limit is defined at a boundary point, but not continuous there.

Finally, we proceed with the proof of the theorem. To simplify notation, we prove it for series with radius of convergence $1$, and leave the simple rescaling to series converging for $|x|<r$ as an exercise. That is, we aim to prove

Proposition 17.1 (Abel’s Theorem for $r=1$) Let $\sum _{k\ge }0a_k{x}^k$ be a power series which converges for $|x|<1$ and assume $\sum _{k\ge 0}{a}_k$ also converges. Then $$\underset{x\to 1^{-}}{lim}\sum _{k\ge 0}{a}_k{x}^k=\sum _{k\ge 0}{a}_k{x}^k$$

Throughout the proof, it is useful to introduce the notation $A_N=\sum _{k=0}^N{a}_k$ and $A=lim{A}_N=\sum _{k\ge 0}{a}_k$ for the partial sums of of the coefficients and their limit. We begin with a lemma providing a means of rewriting the series within the radius of convergence.

Lemma 17.1 Let $\sum a_k{x}^k$ be a power series which converges for $|x|<1$, and $A_N=\sum _{k\le N}{a}_k$ be the partial sums of the coefficients. Then for any $x\in (-1,1)$ $$\sum _{k\ge 0}{a}_k{x}^k=(1-x)\sum _{k\ge 0}{A}_k{x}^k$$

Proof. Recall the formulation of summation by parts given in Abel’s lemma: $$\sum _{k=1}^N{x}_k{y}_k=X_N{y}_N-X_0{y}_0-\sum _{k=0}^{N-1}{X}_k(y_{k+1}-y_k)$$ for $X_N$ the partial sums of the sequence $\{x_k\}$. We can interpret a power series $\sum a_k{x}^k$ as summing a product of two sequences: the sequence $a_k$ of coefficients and the sequence $x^k$ of monomials. Looking at a partial sum of our power series and summing by parts

$$\begin{array}{rl}\sum _{k=0}^N{a}_k{x}^k& =a_0+\sum _{k=1}^N{a}_k{x}^k\\ & =a_0+(A_N{x}^N-A_0{x}^0-\sum _{k=1}^{N-1}{A}_k(x^{k+1}-x^k))\end{array}$$

Using that $A_0=a_0$ and $x^0=1$, this simplifies

$$\begin{array}{rl}& =A_N{x}^N-\sum _{k=1}^{N-1}{A}_k{x}^k(x-1)\\ & =A_N{x}^N+(1-x)\sum _{k=1}^{N-1}{A}_k{x}^k\end{array}$$

As we assumed $A=\sum _{k\ge 0}{a}_k$ converges, $A_N\to A$ as $N\to \mathrm{\infty }$. And as $|x|<1$, we know $x^N\to 0$ by our basic limits. Thus by the limit law for products, $A_N{x}^N\to 0$, so for the full power series

$$\begin{array}{rl}\sum _{k\ge 0}& =\underset{N}{lim}\sum _{k=0}^N{a}_k{x}^k\\ & =\underset{N}{lim}(A_N{x}^N+(1-x)\sum _{k=1}^{N-1}{A}_k{x}^k)\\ & =0+(1-x)\underset{N}{lim}\sum _{k=0}^{N-1}{A}_k{x}^k\\ & =(1-x)\sum _{k\ge 0}{A}_k{x}^k\end{array}$$

Next, we use this new form of the series to convert our problem to something simpler

Lemma 17.2 Let $f(x)=\sum a_n{x}^n$ be a power series with radius of convergence $1$, which also converges at $x=1$. Then $f$ is continuous at $1$ if and only if $$(1-x)\sum _{n\ge 0}(A_n-A)x^n\to 0$$ as $x\to 1^{-}$ for $A_N=\sum _{n\le N}{a}_n$ and $A=\sum _{n\ge 0}{a}_n$.

Proof. The overall goal is to show $$\underset{x\to 1^{-}}{lim}\sum _{k\ge 0}{a}_k{x}^k=\sum _{k\ge 0}\underset{x\to 1^{-}}{lim}{a}_k{x}^k=\sum _{k\ge 0}{a}_k=A$$ Subtracting $A$ from both sides, we must show $\underset{x\to 1^{-}}{lim}\sum _{k\ge 0}{a}_k{x}^k-A=0$. Switching out the series for its alternative expression derived in the previous lemma, our problem is equivalent to showing

$$\underset{x\to 1^{-}}{lim}((1-x)\sum _{k\ge 0}{A}_k{x}^k)-A=0$$

To turn this into something useful, we do a sneaky trick. Recall that for $|x|<1$ we know $\sum _{k\ge 0}{x}^k=\frac{1}{1-x}$ (the geometric series). Clearing the denominator, this means $(1-x)\sum _{k\ge 0}{x}^k=1$, so we “multiply by 1”

$$\begin{array}{rl}(1-x)\sum _{k\ge 0}{A}_k{x}^k-A& =(1-x)\sum _{k\ge 0}{A}_k{x}^k-A\cdot 1\\ & =(1-x)\sum _{k\ge 0}{A}_k{x}^k-A(1-x)\sum _{k\ge 0}{x}^k\\ & =(1-x)(\sum _{k\ge 0}{A}_k{x}^k-\sum _{k\ge 0}A{x}^k)\end{array}$$

Because both of the sums involved are convergent, we can add them term-by-term without changing the value (Proposition 12.1), combining the sums.

$$\begin{array}{rl}& =(1-x)\sum _{k\ge 0}(A_k{x}^k-A{x}^k)\\ & =(1-x)\sum _{k\ge 0}(A_k-A)x^k\end{array}$$

This is just a rewriting of our original series minus the proposed limit. So proving this converges to zero is logically equivalent to our desired result

Finally, we prove Abel’s theorem by showing this does indeed limit to zero, as $x\to 1^{-}$.

Proof. We work directly with the limit definition: for arbitrary $\epsilon >0$ we must provide a $\delta >0$ such that if $1-\delta <x<1$, our sum is less than $\epsilon$ in absolute value. By the triangle inequality and limit inequalities,

$$|(1-x)\sum _{k\ge 0}(A_k-A)x^k|\le |1-x|\sum _{k\ge 0}|A_k-A||x{|}^k=(1-x)\sum _{k\ge 0}|A_k-A|x^k$$

Where the final equality holds as we are concerned with $\underset{x\to 1^{-}}{lim}$ so we can without loss of generality assume $x>0$. So, our goal is to show the right hand side is smaller than $\epsilon$, when $x$ is sufficiently close to 1. Since $A_k\to A$ by definition, there is an $N$ such that for all $k\ge N$, $|A_k-A|<\epsilon$. So, we break our sum into two terms, to estimate separately:

$$(1-x)\sum _{k\ge 0}|A_k-A|x^k=(1-x)\sum _{0\le k<N}|A_k-A|x^k+(1-x)\sum _{k\ge N}|A_k-A|x^k$$

In the second of these sums, we know that $|A_k-A|<\epsilon$, so $$\sum _{k\ge N}|A_k-A|x^k\le \sum _{k\ge N}\epsilon x^k=\epsilon \sum _{k\ge N}{x}^k\le \epsilon \sum _{k\ge 0}{x}^k=\epsilon \frac{1}{1-x}$$

Substituting back into the full second term, $$(1-x)\sum _{k\ge N}|A_k-A|x^k\le (1-x)\frac{\epsilon }{1-x}=\epsilon$$

Thus the second term can be made as small as we like (independently of the value of $x$!), and we need only think about the first term. But this is actually easy: its a finite sum! For any positive $x<1$ we know $x^k<1^k=1$, so

$$\sum _{0\le k<N}|A_k-A|x^k<\sum _{0\le k<N}|A_k-A|$$

Call the value of this finite sum on the right $L$. Then to make the term $(1-x)\sum _{0\le k<N}|A_k-A|x^k$ less than $\epsilon$ it suffices to make $1-x$ less than $\epsilon /L$. So, set $\delta =\epsilon /L$: then for any $x<1$ with $|x-1|<\delta$, we know

$$(1-x)\sum _{0\le k<N}|A_k-A|x^k<(1-x)\sum _{0\le k<N}|A_k-A|=(1-x)L<\frac{\epsilon }{L}L=\epsilon$$

Putting it all back together, we see that for this $\delta$, $|x-1|<\delta$ implies our sum is less than $2\epsilon$. So, we need to go back and replace some epsilons with $\epsilon /2$’s to complete the proof [Apologies: due to the length of this argument, I prioritized readability, and reduced clutter by not writing the correct $\epsilon /2$’s everywhere].

As a last step, we do the substitutions to return from $r=1$ to general radius of convergence.

Exercise 17.3 (The General Case)  

• Let $f(x)=\sum _{n\ge 0}{a}_n{x}^n$ be a power series which converges on $(-r,r)$ and also at $r$. Prove that it is continuous at $r$. *Hint: consider $f(rx)$: show this has radius of convergence $1$ and converges at $1$. Then apply the previous theorem, and re-arrange to yield the result.

• Let $f(x)=\sum _{n\ge 0}{a}_n{x}^n$ be a power series which converges on $(-r,r)$ and also at $-r$. Prove that it is continuous at $-r$. *Hint: consider $f(-x)$.

17.3 $⧫$ Uniqueness

Using this continuity result, we prove a theorem which is very helpful for not getting lost in the world of power series. Its natural to wonder if two distinct power series could converge to the same function (as limits of their partial sums). Of course if any of their coefficients differed no finite partial sums could be equal (as the finite sums are polynomials, and polynomials are fully determined by their coefficients). But this doesn’t rule out any coincidence in the limit. After all, this happens with numbers all the time! Finite decimals are determined by their digits, but infinite decimals are not! $1.00000\dots =0.999999\dots$, and $0.5000=0.4999\dots$, etc.

But this cannot happen for power series: if a function $f(x)$ can be written as a power series in two ways $\sum a_k{x}^k$ and $\sum b_k{x}^k$, then these two are precisely equal: $a_k=b_k$ for all $k$. But we can get by with much less information than this: if two power series agree on a single sequence converging to zero thats already enough information to completely determine them! This is the content of the theorem below.

Theorem 17.3 Let $f(x)=\sum _{k\ge 0}{a}_k{x}^k$ and $g(x)=\sum _{k\ge 0}{b}_k{x}^k$ be two power series with positive radii of convergence (not a priori assumed to be the same). Then if for some sequence $c_n$ of nonzero terms converging to zero we have $f(c_n)=g(c_n)$, it follows that $f$ and $g$ are the same power series: $a_k=b_k$ for all $k$.

Proof. We proceed step by step, starting with the constant terms.
Since $f(x)$ is continuous we can compute $f(0)$ as $limf(c_n)$. Similarly, the continuity of $g$ lets us write $g(0)=limg(c_n)$. But by our assumption that $f(c_n)=g(c_n)$, this implies $f(0)=g(0)$. Plugging in zero to our power series shows $$f(0)=a_0+a_{1}0+a_2{0}^2+\cdots =a_{0}g(0)=b_0+b_{1}0+b_2{0}^2+\cdots =b_0$$ Thus $a_0=b_0$, so the first two terms of our power series are equal.

This argument was remarkably efficient, so let’s try to repeat it. Subtracting the constant terms from $f$ and $g$ yields functions divisible by $x$, and dividing by that $x$ gives $$f_1(x)=\frac{f(x)-a_0}{x}=a_1+a_{2}x+a_3{x}^2+\cdots$$ $$g_1(x)=\frac{g(x)-b_0}{x}=b_1+b_{2}x+b_3{x}^2+\cdots$$

And, as $f_1$ and $g_1$ are again power series with the same coefficients, they have the same radius of convergence (exercise!) and so are continuous at $0$. This means we can re-run our trick:

$$lim{f}_1(c_n)=f_1(lim{c}_n)=f_1(0)=a_1$$ $$lim{g}_1(c_n)=g_1(lim{c}_n)=g_1(0)=b_1$$

But now think a bit about $f_1$ and $g_1$: because $a_0=b_0$, we have actually done exactly the same operation to each function: subtracted the same number and then divided by $x$. Thus, the fact that $f(c_n)=g(c_n)$ immediately implies that $f_1(c_n)=g_1(c_n)$. So, the two sequences we are taking the limit of are the same, meaning their limits are the same: $a_1=b_1$.

You can imagine how we continue from here: induction!

Corollary 17.1 Let $f(x)=\sum _{k\ge 0}{a}_k{x}^k$ and $g(x)=\sum _{k\ge 0}{b}_k{x}^k$ be two power series, which are equal on some neighborhood of zero. Then they are identical, and $a_k=b_k$ for all $k$.

Proof. Let $\epsilon >0$ be some small number where $f(x)=g(x)$ for all $|x|<\epsilon$. Take the sequence $1/n$ and truncate the first finitely many terms until $1/n<\epsilon$, producing a nonzero sequence converging to zero fully contained in $(-\epsilon ,\epsilon )$. Then by our assumption $f$ and $g$ agree on this sequence, so they are equal by the Identity Theorem.