Chapter 10.5, Problem 40E

a.

To determine

To find:The interval of convergence of given series.

Answer to Problem 40E

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ converges in interval $-1<x<1$ or $x\in \left(-1,1\right)$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ .

Formula/ concept used:

Ratio Test: Suppose ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ be a given series and let $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ , then if $L<1$ , the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ is convergent (absolutely convergent) and if $L>1$ the series is divergent.

Calculations:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }{a}_n}={\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$.........(1)

Now, $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{\left(n+1\right)x^{n+1}}{(n+1)+2}}{\frac{nx}{n+2}}\right|$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left|\frac{\left(n+1\right)x^{n+1}}{(n+1)+2}\times \left(\frac{n+2}{n{x}^n}\right)\right|$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left|\left(\frac{n+1}{n}\right)\times \left(\frac{n+2}{n+3}\right)\frac{x^{n+1}}{x^n}\right|$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left|\left(1+\frac{1}{n}\right)\times \left(\frac{1+\frac{2}{n}}{1+\frac{3}{n}}\right)\frac{x^{n+1}}{x^n}\right|$

  $\Rightarrow L=\left|\left(1+0\right)\times \left(\frac{1+0}{1+0}\right)x\right|$$\because \underset{n\to \infty }{\lim }\frac{a}{n}=0$

  $\Rightarrow L=\left|x\right|$

The series (1) is convergent if $L=1\Rightarrow \left|x\right|<1$

  $\Rightarrow -1<x<1$

Thus, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ converges in interval $-1<x<1$ or $x\in \left(-1,1\right)$ .

Conclusion:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ converges in interval $-1<x<1$ or $x\in \left(-1,1\right)$ .

b.

To determine

To find:The value of x for which the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ is absolutely convergent.

Answer to Problem 40E

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ is absolutely convergent in interval $-1<x<1$ .

Explanation of Solution

Given:

Theseries ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ in part (a).

Concept used:

Suppose ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ be a given series and let $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ , then if $L<1$ , the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ is convergent (absolutely convergent) and if $L>1$ the series is divergent.

Calculations:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ .

As shown in part (a) $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|x\right|<1$ .

Therefore, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ is absolutely convergent in interval $-1<x<1$ .

Conclusion:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ is absolutely convergent in interval $-1<x<1$ .

c.

To determine

To find: The value of x for which the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ is conditionally convergent.

Answer to Problem 40E

There is no interval in which the series converge conditionally.

Explanation of Solution

Given:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ in part (a).

Concept used:

The series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ is conditionally convergent if the series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ converges but ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ diverges.

If the original series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ is absolutely convergent then series does not converge conditionally.

If a series is alternating and not absolutely convergent, we check for conditional convergence using the alternating series test.

Calculations:

As shown in part (a) and (b) the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ and $\left|{\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}\right|$ both converge in the interval $-1<x<1$ .

Since, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ absolutely convergentin interval, $-1<x<1$ , therefore there is no interval in which the series converge conditionally.

Conclusion:

There is no interval in which the series converge conditionally.