$$$$

20  Power Series

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

Definition 20.1 (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 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.

The simplest power series are polynomials themselves, which have $a_n=0$ 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 $(-1,1)$.

The domain of a general power series will always look like an interval, motivating the definition of a radius of convergence.

Definition 20.2 The radius of convergence of a power series is the largest $r>0$ such that the series converges at each point of $(-r,r)$.

General power series are essentially just slight modifications the geometric series, multiplying each power of $x$ by some coefficient. Thus, its natural to seek their radius of convergence by comparison with geometric series (the Ratio test):

Theorem 20.1 (Power Series and Radius of Convergence) If $\sum _k{a}_k{x}^k$ is a power series, let $\alpha =lim|a_{k+1}/a_k|$. If $\alpha =0$ then the power series converges on the entire real line. And, if $\alpha \ne 0$ then its radius of convergence is $R=1/\alpha$: the series converges in $(-R,R)$ and diverges for $|x|>R$.

Proof. Choose $x\in \mathbb{R}$ and compute the ratio test for the corresponding series $$lim\left|\frac{a_{k+1}{x}^{k+1}}{a_k{x}^k}\right|=|x|lim\left|\frac{a_{k+1}}{a_k}\right|=|x|\alpha$$

If $\alpha =0$ then this entire quantity is zero, independent of $x$. And, as $0<1$ the ratio test ensures the series converges. For $\alpha \ne 0$, the ratio test gives convergence if $|x|\alpha <1$, or $|x|<1/\alpha =R$, and divergence for $|x|>R$, as claimed.

Inside their radius of convergence power series are always very well behaved:

Corollary 20.1 Power series converge absolutely within their radius of convergence.

Proof. If $x$ is within the radius of convergence, then the ratio test on $\sum _k{a}_k{x}^k$ yields a number strictly less than $1$ by definition, signaling absolute convergence.

Since it can often be difficult to determine exactly what happens at the endpoints of the interval of convergence, where the series may converge either absolutely, conditionally, or not at all. Thus speaking of the radius (and avoiding the issue of convergence at endpoints) is often useful.

20.1 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 20.1 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 20.1 Show the power series $\sum \frac{x^n}{n}$ has domain $[-1,1)$.

Exercise 20.2 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 20.3 Show the power series $\sum n!x^n$ diverges for all $x\ne 0$.

Exercise 20.4 Series $\sum 2^n{x}^n$ converges on $[-1/2,1/2)$. Hint: substitution $y=2x$

Example 20.2 Where does $\sum 2^n{x}^{3n}$ converge? Trickier! Need to worry about the exponents not being just $n$

20.2 Power Series for Functions

In the above section we discovered many new functions: its easy to define functions that no one has ever heard of by writing down new power series! And indeed, this is often how new functions are first described.

But another big use of power series will be to provide formulas for functions we already know about. At the moment we have essentially one function that we know a power series for: the geometric series $$\sum _{k\ge 0}{x}^k=\frac{1}{1-x}$$

From this we can build many new functions, via substitution:

Example 20.3 A series for $\frac{1}{1+x}$ can be constructed, by substituting $-x$ for $x$ on both sides of the equality above:

$$\begin{array}{rl}\frac{1}{1+x}& =\frac{1}{1-(-x)}\\ & =\sum _{k\ge 0}(-x{)}^k\\ & =\sum _{k\ge 0}(-1{)}^k{x}^k\\ & =1-x+x^2-x^3+x^4-x^5+\cdots \end{array}$$

Its instructive to try this with a couple examples yourself.

Exercise 20.5 A series for $\frac{1}{1+x^2}$

Exercise 20.6 A series for $\frac{x}{2+3x^2}$

In the study of calculus, we will produce much more powerful tools to find power series of functions.