Power series

In this post, we deal with a class of series called the power series that contains a variable.

Definition (Power series)

Let $\{a_n\}$ be a sequence of real numbers, $b$ a real number, and $x$ a variable $x$. The series given by

$$\begin{array}{}\text{(Eq:PS)}& \sum _{n=0}^{\mathrm{\infty }}{a}_n(x-b{)}^n=a_0+a_1(x-b)+a_2(x-b{)}^2+\cdots \end{array}$$

is called a power series centered at $x=b$.

If we set $t=x-b$ in (Eq:PS), we obtain a power series centered at $t=0$. In most practical cases, it suffices to deal with power series centered at $x=0$.

Example. A polynomial of $x$, $f(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$ can be regarded as a power series by setting $a_{n+1}=a_{n+2}=\cdots =0$.

In general, the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ is a polynomial of $x$ if $a_n=0$ for all but finitely many $n$. □

If the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ is a polynomial, we can substitute an arbitrary real number to $x$ to calculate the sum. If it is not a polynomial (i.e., $a_n\ne 0$ for infinitely many $n$), then substituting real numbers to $x$ other than 0 may result in the divergence of the power series. As long as the power series converges (i.e., has a sum), we may regard it as a function of $x$.

Radius of convergence

Theorem (Convergence of power series)

If the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges at $x=u(\ne 0)$, then it converges absolutely for all $x$ such that $|x|<|u|$.

Proof. Since $\sum _{n=0}^{\mathrm{\infty }}{a}_n{u}^n$ has a sum, the sequence $\{a_n{u}^n\}$ converges to 0. In particular, $\{a_n{u}^n\}$ is bounded. Therefore, there exists an $M>0$ such that $|a_n{u}^n|<M$ for all $n\ge 0$. For any $x\in \mathbb{R}$, let us define $r=\left|\frac{x}{u}\right|$. We have

$$|a_n{x}^n|=|a_n{u}^n\cdot \frac{x^n}{u^n}|\le M{r}^n.$$

If $|x|<|u|$, then $r<1$. Therefore, the series $\sum _{n=0}^{\mathrm{\infty }}|a_n{x}^n|$ has a dominating series $\sum _{n=0}^{\mathrm{\infty }}M{r}^n$. It follows that $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges absolutely if $|x|<|u|$. ■

Definition (Radius of convergence)

Given a power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$, the quantity defined by

$$r=sup\left\{|u|\right|\sum _{n=0}^{\mathrm{\infty }}{a}_n{u}^n\text{ converges}\}$$

is called the radius of convergence of the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$.

Clearly, the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges if $x=0$. Therefore $r\ge 0$. It is possible that $r=+\mathrm{\infty }$.

From the above theorem (convergence of power series), we can see the following ($r$ is the radius of convergence):

• If $0<r<+\mathrm{\infty }$, the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges absolutely at all $x$ such that $|x|<r$, and diverges at all $x$ such that $|x|>r$.
• If $r=+\mathrm{\infty }$, the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges at all $x\in \mathbb{R}$.
• If $r=0$, the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ diverges at all $x\ne 0$.

In particular, if $r>0$, we can regard $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ as a function of $x$ on the open interval $(-r,r)$.

Example. The power series $\sum _{n=0}^{\mathrm{\infty }}{x}^n$ converges for $|x|<1$ and diverges for $|x|>1$. Therefore, its radius of convergence is 1. □

Remark. When the radius of convergence of the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ is $r$, we cannot generally decide whether the series converges or not at $|x|=r$. □

Calculating the radius of convergence

From the Cauchy and D'Alembert criteria, we have the following theorem:

Theorem (Radius of convergence)

For the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ with its radius of convergence $r$, the following hold:

1. If $l=\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}$, then $r=\frac{1}{l}$.
2. If $l=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|$, then $r=\frac{1}{l}$.

Remark. In this theorem, we include the case where the limit does not exist, i.e., $l=+\mathrm{\infty }$, and formally define $\frac{1}{0}=+\mathrm{\infty }$ and $\frac{1}{\mathrm{\infty }}=0$. □

Proof. Exercise. ■

See also: Convergence of series for Cauchy and D'Alembert's criteria.

Example. Let us find the radius of convergence of $\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$.

For $a_n=\frac{1}{n!}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n+1}=0.$$

Thus, the radius of convergence is $+\mathrm{\infty }$. □

Example. Consider the following series:

$$\begin{array}{}\text{(Eq:x2n)}& \sum _{n=0}^{\mathrm{\infty }}{x}^{2n}=1+0\cdot x+x^2+0\cdot x^3+x^4+\cdots .\end{array}$$

The coefficient $a_n$ is either 1 or 0. Thus, $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}$ does not exist, and the sequence $\left\{\frac{a_{n+1}}{a_n}\right\}$ cannot be defined. Therefore, we cannot use the above theorem to find the radius of convergence for this series.  □

However, if we use the limit superior, we can always write the radius of convergence as

$$r=\frac{1}{\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|a_n|}}.$$

See also: Limit superior, limit inferior.

Example. The series in (Eq:x2n), 

$$\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|a_n|}=1.$$

Therefore, the radius of convergence is 1. □