Question

In Problems 1-10 find the interval and radius of convergence for the given power series.

$\sum _{n=1}^{\infty }\frac{1}{n^2}{x}^n$

Short answer

The radius of convergence is$R=1,\text{interval:}[-1,1]$

Step-by-step solution

Ratio test

The Ratio test for the power series,$\sum _{n=1}^{\infty }{a}_n{(x-a)}^n$is given as,

$\underset{n\to \infty }{lim}\left|\frac{{\displaystyle a_{n+1}{(x-a)}^{n+1}}}{{\displaystyle a_n{(x-a)}^n}}\right|=L$

If$L<1$, then the series converges absolutely.

If, $L>1$then the series diverges.

If, L=1 then the test is inconclusive.

Apply ratio test

Consider$\sum _{n=1}^{\infty }\frac{1}{n^2}{x}^n$

Let$c_n=\frac{x^n}{n^2}$then$c_{n+1}=\frac{x^{n+1}}{{(n+1)}^2}$

The ratio test gives,

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{\frac{x^{n+1}}{{(n+1)}^2}}{\frac{x^n}{n^2}}\right|=\underset{n\to \infty }{\lim }\left|\frac{x^{n+1}}{{(n+1)}^2}\cdot \frac{n^2}{x^n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{x^n\cdot x}{{(n+1)}^2}\cdot \frac{n^2}{x^n}\right|(\text{since}{a}^m\cdot a^n=a^{m+n})\\ =\underset{n\to \infty }{\lim }\left|\frac{x^n\cdot x}{{(n+1)}^2}\cdot \frac{n^2}{x^n}\right|\\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{n^2}{n^2{(1+\frac{1}{n})}^2}\right|\\ =\left|x\right|\end{array}$

Radius & interval of convergence

To determine the interval and radius of convergence:

For convergence,$\left|x\right|<1$

This inequality is the power series is centered at.$a=0$

Thus the radius and the interval of convergence are$R=1,(-1,1)$

Find convergence at end points

For our problem, the endpoints are$x=-1$and.$x=1$

At, $x=-1$the series becomes$\sum _{n=1}^{\infty }\frac{1}{n^2}{(-1)}^n=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}$

Since the term$\frac{1}{n^2}$is non-negative and the limit approaches 0 as, $n\to \infty$the series is convergent by the alternating series test.

At,$x=1$ the series becomes$\sum _{n=1}^{\infty }\frac{{\left(1\right)}^n}{n^2}=\sum _{n=1}^{\infty }\frac{1}{n^2}$

This series is convergent by the p-series test since.$2>1$

So, the power series is convergent at both endpoints.

So the radius and interval of convergence of the above power series are.$R=1,\text{interval:}[-1,1]$