Chapter 8, Problem 30RQ

To determine

To Prove: The radius of the convergence and interval of the convergence of the series ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{x^n}{5^n}$ .

Answer to Problem 30RQ

The interval of convergence is $\left[-5,5\right]$ and the radius of convergence is $R=5$ .

Explanation of Solution

Calculation:

Consider the given series is ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{x^n}{5^n}$ .

Consider that,

  $\begin{array}{c}{a}_n=\left|\frac{{\left(-1\right)}^{n+1}}{{\left(n+1\right)}^2}\frac{x^{n+1}}{5^{n+1}}\cdot \frac{n^2{5}^n}{{\left(-1\right)}^n{x}^n}\right|\\ =\left|-x\frac{n^2}{5{\left(n+1\right)}^2}\right|\\ =\frac{1}{5}\frac{1}{{\left(1+\frac{1}{n}\right)}^2}\left|x\right|\end{array}$

Consider the limit $n\to \infty$ then,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}\underset{n\to \infty }{\lim }\frac{1}{{\left(1+\frac{1}{n}\right)}^2}\left|x\right|\\ =\frac{1}{5}\left|x\right|\end{array}$

Then, by the ratio test consider that series is convergent if $\frac{1}{5}\left|x\right|<1$ .

Then, $\left|x\right|<5$ and the diverges when $\frac{1}{5}\left|x\right|>1$ for $\left|x\right|>5$ . This suggest that the radius of the convergence is $R=5$ and $-5<x<5$ .

Consider that $x=5$ . Then, the series becomes ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{5^n}{n^2{5}^n}={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{1}{n^2}$ . Then, the alternating series with $b_n=\frac{1}{n^2}$ then, $b_{n+1}\le b_n$ and $\underset{n\to \infty }{\text{lim}}\frac{1}{n^2}=0$ . This, shows that the series is convergent.

For $x=-5$ . Then, ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n{\left(-5\right)}^n}{n^2{5}^n}}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ . Consider that p series with the $p=2>1$ so that it is convergent and the interval of convergence is $\left[-5,5\right]$ .