$$$$

14  Power Series

Highlights of this Chapter: we introduce the definition of a power series, and testing for convergence via ratios.

We have developed some pretty powerful tools to prove the convergence of infinite series: for example, with little work we can apply the ratio test to immediately see $\sum \frac{n}{2^n}$ converges.

Indeed the same technique shows that the more general series $\sum \frac{n{\alpha }^n}{2^n}$ converges whenever $|\alpha |<2$. This invites a bit of a change in perspective: we might think of the series above as a function that takes in one value of $\alpha$ (a parameter), and outputs the limit of the series. From this perspective, our calculation above actually helped us find the domain of the function - all the values of $\alpha$ that make sense to plug in.

Defining functions using sequences and series proves to be an incredibly powerful tool in analysis, and so we take a break from our more theoretical development to introduce the simplest and most useful case: power series. Polynomials or finite sums of multiples of powers of $x$, are some of the simplest functions we know. So its natural to wonder about their infinite counterparts: power series, arising as the limit of a sequence of polynomials of increasing degree

Definition 14.1 (Power Series) A power series is a function defined as the limit of a sequence of polynomials $$f(x)=\sum _{n\ge 0}{a}_n{x}^n$$ for a sequence $a_n$ of real numbers. For each $x$, this defines an infinite series; the domain of a power series is the subset $D\subset \mathbb{R}$ of $x$ values where the series converges.

Of course, polynomials themselves are a special case of power series, with $a_n=0$ after some finite $N$. Perhaps the second simplest power series is the one with $a_n=1$ for all $n$:

$$f(x)=1+x+x^2+x^3+x^4+\cdots +x^n+\cdots$$

This is none other than the geometric series in $x$! So, it converges whenever the common ratio $x$ satisfies $|x|<1$: its domain is the interval $(-1,1)$.

Power series are an extremely versatile tool to reach beyond the arithmetic of polynomials, while staying close to the fundamental operations of addition/subtraction and multiplication/division. One of our main uses of them will be to provide efficient means of computing important functions (exponentials, logs, trigonometric functions, etc).

Definition 14.2 (Power Series Representation) Given a function $f(x)$, a power series representation of $f$ is a series $s(x)=\sum a_n{x}^n$ such that $s(x)=f(x)$ whenever $s(x)$ converges.

Example 14.1 The function $f(x)=\frac{1}{1-x}$ has a power series representation on the interval $(-1,1)$, where $$\sum _{n\ge 0}{x}^n=\frac{1}{1-x}$$

This might not seem like an exciting discovery, as $1/(1-x)$ is a function that is easy for us to compute using the field operations, and now we’ve build a more complicated looking expression - an infinite series - to compute the same value! The ability to represent a function as a power series will be much more useful when looking at functions that are difficult to understand, like the exponential. Our best procedure for computing an irrational power right now is the definition: the limit of a sequence of rational powers. But such a limit is ridiculously hard to compute in practice. If we could instead represent the exponential as a power series, we could replace this limit with a series made out of just addition and multiplication! We will do exactly this, in a future chapter.

14.1 Convergence

Here we will study the most general theory of power series $\sum a_n{x}^n$ for arbitrary sequences $a_n$. The first thing to understand is their domain: for which values of $x$ do the series converge? One point stands out immediately: for $x=0$ the terms $a_n{x}^n$ are all equal to zero, so the partial sums of $\sum a_n{x}^n$ are $0+0+0\cdots +0=0$, and so the series converges to $0$. Thus, $x=0$ is in the interval of convergence of every power series.

Proposition 14.1 If a power series $\sum a_n{x}^n$ converges at some $u>0$, then it converges at all $x\in (-u,u)$.

Proof. If $\sum a_n{u}^n$ converges, then by the root test $lim sup\sqrt[n]{|a_n{u}^n|}\le 1$ (by process of elimination: if this sequence were unbounded or has limsup $>1$ then we know it diverges). Now $x\in (-u,u)$ means $|x|<|u|$. Noting that $x=(x/u)u$, we may rewrite the limit we want as follows:

$$\sqrt[n]{|a_n{x}^n|}=|x|\sqrt{n}|a_n|=\frac{|x|}{|u|}|u|\sqrt[n]{|a_n|}=\frac{|x|}{|u|}\sqrt[n]{|a_n{u}^n|}$$

Because $|x|/|u|$ is a positive constant, we can pull it out of the limsup and compute

$$lim sup\frac{|x|}{|u|}\sqrt[n]{|a_n{u}^n|}=\frac{|x|}{|u|}lim sup\sqrt[n]{|a_n{u}^n|}$$

This is strictly less than $1$ as $|x|/|u|$ is strictly less than 1 (since $|x|<|u|$) and the limsup term is less than or equal to 1 by our assumption of convergence. Thus, with limit less than 1, this series converges absolutely by the root test, so $\sum a_n{x}^n$ converges as claimed.

This motivates the following definition:

Definition 14.3 (Radius of Convergence) The radius of convergence of a power series is the largest value of $r>0$ such that the series converges on $(-r,r)$.

With a more careful use of our convergence tests, we can compute an exact expression for the radius of convergence in terms of the coefficients. This will be of incredible theoretical utility in the chapters to come, and is often known as the Cauchy Hadamard Theorem.

Theorem 14.1 (Finding the Radius of Convergence: Cauchy-Hadamard) Let $\sum a_n{x}^n$ be a power series, and $\alpha =lim sup\sqrt[n]{|a_n|}$. Then the radius of convergence is $r=1/\alpha$ (where $\alpha =0$ means convergence on all of $\mathbb{R}$), and if $\sqrt[n]{|a_n|}$ is unbounded (so its $lim sup$ is undefined, or infinite) then $\sum a_n{x}^n$ converges only at $x=0$.

Proof. First we deal with the generic case, where $\alpha$ is a nonzero real number. Since $lim sup\sqrt[n]{|a_n{x}^n|}=|x|lim sup\sqrt[n]{|a_n|}=|x|\alpha$, we know this series converges whenever $|x|\alpha <1$, so $|x|<1/\alpha$. Since the root test further ensures the series diverges whenever $|x|\alpha >1$, $r=1/\alpha$ is the largest number such that the series converges for all $|x|<r$, making it the radius of convergence.

The same reasoning works when $\alpha =0$, as now for any $x$ we see $lim sup\sqrt[n]{|a_n{x}^n|}=|x|lim sup\sqrt[n]{|a_n|}=|x|\alpha =0$, which is less than $1$: thus $\sum a_n{x}^n$ converges for all $x\in \mathbb{R}$.

Finally, if $\sqrt[n]{|a_n|}$ is an unbounded sequence, then for any nonzero $x\ne 0$ the sequence $|x|\sqrt[n]{|a_n|}=\sqrt[n]{|a_n{x}^n|}$ is also unbounded. Taking $n^{th}$ powers shows $|a_n{x}^n|$ is unbounded, and in particular does not converge to zero. Thus $a_n{x}^n$ does not converge to zero, so the series diverges by the divergence test. Since all series converge for $x=0$, this makes $\{0\}$ the entire domain, so the radius of convergence is zero.

Corollary 14.1 (Absolute Convergence of Power Series) Let $f(x)=\sum a_n{x}^n$ be a power series with radius of convergence $r$, and let $u\in (-r,r)$. Then $f$ converges absolutely at $u$.

The root test further implies that for a power series with radius of convergence $r$, at any $x$ with $|x|>|r|$ the series must diverge. Thus, the domain of a power series has a very limited form: it must be an interval centered at zero. Because of this we often call the domain the interval of converence.

Definition 14.4 (Interval of Convergence) The interval of convergence of a power series $\sum a_n{x}^n$ is another word for its domain, the set of all $x$ for which the series converges.

Exercises below show there are no further restrictions: any possible interval centered at zero is the interval of convergence for some power series: so domains can take the form $\{0\},(-r,r)$, $[-r,r]$, $[-r,r)$, $(-r,r]$, or $(-\mathrm{\infty },\mathrm{\infty })$; all of $\mathbb{R}$.

While the root test is of great theoretical utility to proving the above theorem, in practice $n^{th}$ roots are rather unweildly to work with, and we might wish to instead apply hte ratio test. One can make a version of the ratio test specifically adapted to power series as follows:

Exercise 14.1 Let $\sum a_n{x}^n$ be a power series, and assume the sequence of ratios $\frac{a_{n+1}}{a_n}$ converges to some $\alpha \in \mathbb{R}$. Prove that the radius of convergence is $r=1/\alpha$ when $\alpha \ne 0$, and converges on all of $\mathbb{R}$ when $\alpha =0$.

14.1.1 Skipping Terms

The root and ratio tests as proven above apply to series $\sum a_n{x}^n$, where the $n^{th}$ term is an $n^{th}$ power of $x$. But there are very many natural series not of this form: for example, a series with only even powers of $x$,

$$1+\frac{x^2}{4}+\frac{x^4}{16}+\frac{x^6}{64}+\cdots =\sum _{n\ge 0}\frac{x^{2n}}{4^n}$$

Or only odd powers, or only powers that are multiples of three, etc. It might be tempting to directly apply the root test (for example) to the coefficients of such a series and conclude

$$\sqrt[n]{1/4^n}=1/4$$

So the series converges for $|x|<4$. However this would be wrong! A more careful application of the root test to the entire series shows

$$\sqrt[n]{x^{2n}/4^n}=|x^2|/4$$

So we actually need $x^2/4<1$, so $x^2<4$ so $|x|<2$. Being careful and applying the root test / ratio test to the entire series each time (instead of the shortcut where its applied to the coefficients) will avoid trouble in such cases.

14.2 Problems

14.2.1 Example Power Series

Power series provide us a means of describing functions via explicit formulas that we have not been able to thus far, by allowing a limiting process in their definition. For instance, we will soon see that the power series below is an exponential function.

Exercise 14.2 Show the power series $\sum \frac{x^n}{n!}$ converges for all $x\in \mathbb{R}$.

When a power series converges on a finite interval, its behavior at each endpoint may require a different argument than the ratio test (as that will give $1$, and tell you nothing)

Example 14.2 Show the power series $\sum \frac{x^n}{n}$ has domain $[-1,1)$. This shows that its possible for power series to have a domain which is closed on the left side and open on the right side of the interval of convergence.

To finish showing all possible interval types exist, create an example of a power series which converges on an interval of the form $(-a,a]$ for some $a>0$ (and prove your claim).

Exercise 14.3 Show the power series $\sum \frac{x^n}{n^2}$ has domain $[-1,1]$.

When the radius of convergence is $0$, the power series converges at a single point:

Exercise 14.4 Show the power series $\sum n!x^n$ diverges for all $x\ne 0$.

Exercise 14.5 Series $\sum 2^n{x}^n$ converges on $[-1/2,1/2)$. Hint: substitution $y=2x$

Example 14.3 Where does $\sum 2^n{x}^{3n}$ converge? Trickier! Need to worry about the exponents not being just $n$