Chapter 8.5, Problem 3E

To determine

To calculate:The radius of convergence and interval of convergence of the series.

Answer to Problem 3E

The Radius of convergence is $R=1$ and the interval of convergence is $\left[-1,1\right)$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{x^n}{\sqrt{n}}}$

Concept Used:

Ratio Test:

(i) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$is absolutely convergent.

(and therefore convergent)

(ii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$is divergent.

(iii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series is ${\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{x^n}{\sqrt{n}}}$

  $a_n=\frac{x^n}{\sqrt{n}}$

Then,

  $a_{n+1}=\frac{x^{n+1}}{\sqrt{n+}1}$

Use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{x^{n+1}}{\sqrt{n+1}}\times \frac{\sqrt{n}}{x^n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{x\times \sqrt{n}}{\sqrt{n+1}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\sqrt{1}}{\sqrt{1+\frac{1}{n}}}\right|\times \left|x\right|\to \left|x\right|asn\to \infty \end{array}$

By the ratio test, the given series converges if $\left|x\right|<1$ and diverges if $\left|x\right|>1$ .

This means the radius of convergence is $R=1$

And the series converges in the interval $\left(-1,1\right)$ , but for convergence at the end points of this interval.

If $x=-1$ the series become

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{{(-1)}^n}{\sqrt{n}}}\\ =\frac{-1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots \cdots \cdots \cdots \cdots \cdots \end{array}$

Which converges by alternating series test.

If $x=1$ the series become

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{{(1)}^n}{\sqrt{n}}}\\ =\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots \cdots \cdots \cdots \cdots \cdots \end{array}$

Which diverges by Intergral test.

Therefore the given series converges when $-1\le x<1$

So the interval of convergence is $\left[-1,1\right)$

Conclusion:

The Radius of convergence is $R=1$ and the interval of convergence is $\left[-1,1\right)$ .