9.6 Convergence of Power Series
Power Series
A power series about $\boldsymbol{x}=\mathbf{0}$ is an infinite series of the form
$
\sum_{n=0}^{\infty} a_n x^n=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots+a_n x^n+\cdots
$
where $x$ is a variable and the $a_n$ ‘s are constants.
More generally, a series of the form
$
\sum_{n=0}^{\infty} a_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+\cdots+a_n(x-c)^n+\cdots
$
is called a power series centered at $c$, or a power series about $c$.
Convergence of a Power Series
For a power series centered at $c$ there are only three possibilities:
1. The series converges only at $x=c$.
2. The series converges for all $x$.
3. There exists a real number $R>0$ such that the series converges for $|x-c|<R$, and diverges for $|x-c|>R$.
The number $R$ is called the radius of convergence of the power series.
In most cases, $R$ can be found by using the Ratio Test. The Ratio Test always fails when $x$ is an endpoint of the interval of convergence, so each endpoint must be tested separately for convergence or divergence.
If the series converges only at $c$, the radius of convergence is $R=0$. If the series converges for all $x$, the radius of convergence is $R=\infty$. The set of all values of $x$ for which the power series converges is the interval of convergence of the power series.
Example 1
Find the radius of convergence and interval of convergence of the series $\sum_{n=0}^{\infty} \frac{(-2)^n x^n}{\sqrt{n+3}}$
▶️Answer/Explanation
Solution
Let $a_n=(-2)^n x^n / \sqrt{n+3}$.
$
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(-2)^{n+1} x^{n+1}}{\sqrt{n+4}} \cdot \frac{\sqrt{n+3}}{(-2)^n x^n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(-2) x \sqrt{n+3}}{\sqrt{n+4}}\right|=2|x|
$
By the ratio test, the given series converges if $2|x|<1$ or $|x|<\frac{1}{2}$.
Thus the radius of convergence is $R=\frac{1}{2}$.
The inequality $|x|<\frac{1}{2}$ can be written as $-\frac{1}{2}<x<\frac{1}{2}$.
We must now test for convergence at the end points of this interval.
When $x=-\frac{1}{2}$, the series becomes
$
\sum_{n=0}^{\infty} \frac{(-2)^n(-1 / 2)^n}{\sqrt{n+3}}=\sum_{n=0}^{\infty} \frac{1}{\sqrt{n+3}}=\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{6}}+\cdots
$
which is a $p$-series and diverges since $p=1 / 2<1$.
When $x=\frac{1}{2}$, the series becomes
$
\sum_{n=0}^{\infty} \frac{(-2)^n(1 / 2)^n}{\sqrt{n+3}}=\sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+3}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}+\cdots
$
which converges by the Alternating Series Test.
So the interval of convergence is $\left(-\frac{1}{2}, \frac{1}{2}\right]$.
Example 2
Find the radius of convergence and interval of convergence of the series
$
\sum_{n=0}^{\infty} n !(2 x)^n
$
▶️Answer/Explanation
Solution
Let $a_n=\frac{n^2 x^n}{n !}$.
$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(n+1)^2 x^{n+1}}{(n+1) !} \cdot \frac{n !}{n^2 x^n}\right|=\lim _{n \rightarrow \infty}\left|\frac{(n+1)^2 x}{n^2(n+1)}\right| \\
& =|x| \lim _{n \rightarrow \infty}\left|\frac{n+1}{n^2}\right|=0
\end{aligned}
$
The series converges for all $x$, so $R=\infty$ and the interval of convergence is $(-\infty, \infty)$.
Example3
What are all values of $x$ for which the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n^3}$ converges?
(A) $-1<x<1$
(B) $-1 \leq x \leq 1$
(C) $-1<x \leq 1$
(D) $-1 \leq x<1$
▶️Answer/Explanation
Ans:B
Example5
What are all values of $x$ for which the series $\sum_{n=0}^{\infty} \frac{n(x-2)^n}{3^n}$ converges?
(A) $-1<x<5$
(B) $-1<x \leq 5$
(C) $-2 \leq x<4$
(D) $-2<x \leq 4$
▶️Answer/Explanation
Ans:A
Example6
What are all values of $x$ for which the series $\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1) !}$ converges?
(A) $0<x<2$
(B) $0 \leq x<2$
(C) $-1<x \leq 2$
(D) All real $x$
▶️Answer/Explanation
Ans:D