5.4 Power Series

A power series is a series of the powers of some unknown variable, basically an infinite polynomial. We make a clearer definition here.

Definition 5.5.

Let $(a_{n})_{n}$ be a sequence of real numbers. Then a power series in some variable $x$ is given by

\sum_{n=1}^{\infty}a_{n}x^{n}.

(5.26)

If we compare the form of a power series to a geometric series, we can see that the convergence of a power series depends on the value of the variable $x$ , so the power series is a function of $x$ . In particular, it can be shown that a power series either converges for all $x\in\mathbb{R}$ , or it converges in some interval of the real line symmetric about $x=0$ . For example, if $(a_{n})_{n}$ is the constant sequence $(1)_{n}$ , then the power series $\sum_{n=1}^{\infty}x^{n}$ converges for $-1<x<1$ .

Theorem 5.9

Consider a power series $\sum_{n=1}^{\infty}a_{n}x^{n}$ and suppose $\lim_{n\to\infty}\lvert a_{n}\rvert^{\frac{1}{n}}=\beta$ . Then, taking $R=\frac{1}{\beta}$ (if $\beta=0$ , then take $R=\infty$ ; if $\beta=\infty$ , then take $R=0$ ), we have

(i)

The power series converges for $x\in(-R,R)$ .
(ii)

The power series diverges for $x\notin(-R,R)$ .

$R$ is called the radius of convergence of the power series.

Proof.

Fix $x\in\mathbb{R}$ . Then

\lvert a_{n}x^{n}\rvert^{\frac{1}{n}}=\lvert a_{n}\rvert^{\frac{1}{n}}\lvert x% \rvert\to\beta\lvert x\rvert\text{ as }n\to\infty.

(5.27)

(i)
1. (a)
  
  $\beta=0$ (so $R=\infty$ and $\lvert x\rvert<R$ ) implies $\beta\lvert x\rvert<1$ , so by the root test $\sum_{n=1}^{\infty}a_{n}x^{n}$ converges.
2. (b)
  
  $\beta>0$ and $\beta\lvert x\rvert<1$ (so $\lvert x\rvert<R$ ). By the root test $\sum_{n=1}^{\infty}a_{n}x^{n}$ converges.
(ii)
1. (a)
  
  $\beta=\infty$ (so $R=0$ ). Then the root test implies that $\sum_{n=1}^{\infty}a_{n}x^{n}$ diverges.
2. (b)
  
  $\beta>0$ and $\beta\lvert x\rvert>1$ (so $\lvert x\rvert>R$ ). Then by the root test $\sum_{n=1}^{\infty}a_{n}x^{n}$ diverges.

∎

Note that when we are finding $R$ , it can often be much easier to find the limit $\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_{n}}\right\rvert$ than $\lim_{n\to\infty}\lvert a_{n}\rvert^{\frac{1}{n}}$ , and theorem 5.8 (iii) says that they have the same value.

Example 5.8.

For what values of $x$ does the power series $\sum_{n=1}^{\infty}\frac{n+1}{2^{n}}x^{n}$ converge?
Using the root test, we see that

\lvert a_{n}\rvert^{\frac{1}{n}}=\left(\frac{n+1}{2^{n}}\right)^{\frac{1}{n}}=% \frac{(n+1)^{\frac{1}{n}}}{2}\to\frac{1}{2}\text{ as }n\to\infty.

(5.28)

So the radius of convergence of this power series is $R=2$ . What happens for the bounding values? For $x=2$ , the power series becomes $\sum_{n=1}^{\infty}\frac{n+1}{2^{n}}2^{n}=\sum_{n=1}^{\infty}(n+1)$ which is divergent. For $x=-2$ , the power series is $\sum_{n=1}^{\infty}\frac{n+1}{2^{n}}(-2)^{n}=\sum_{n=1}^{\infty}(-1)^{n}(n+1)$ which is divergent. Hence the power series converges $\forall x\in(-2,2)$ .

Example 5.9.

For what values of $x$ does the power series $\sum_{n=1}^{\infty}\frac{1}{n!}x^{n}$ converge?
Inspecting the ratio of consecutive terms, we see

\frac{\lvert a_{n+1}\rvert}{\lvert a_{n}\rvert}=\frac{\frac{1}{(n+1)!}}{\frac{% 1}{n!}}=\frac{n!}{(n+1)!}=\frac{1}{n+1}\to 0.

(5.29)

Hence $\beta=0$ and $R=\infty$ , so the power series converges $\forall x\in\mathbb{R}$ .