Chapter 10.5, Problem 39E

a.

To determine

To find:The interval of convergence of given series.

Answer to Problem 39E

The interval of convergence of series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is $-8<x<12$ or $x\in \left(-8,12\right)$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ .

Formula/ concept used:

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

If the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ is convergent then ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ is convergent.

Calculations:

The given series is ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$.........(1)

Now, $L=\underset{n\to \infty }{\lim }|a_n{|}^{1/n}=\underset{n\to \infty }{\lim }{\left|\frac{{\left(x-2\right)}^n}{{10}^n}\right|}^{1/n}$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left|\frac{\left(x-2\right)}{10}\right|=\left|\frac{\left(x-2\right)}{10}\right|$

The series (1) is convergent if $L=\left|\frac{\left(x-2\right)}{10}\right|<1$

  $\Rightarrow \left|\left(x-2\right)\right|<10$

  $\Rightarrow -10<\left(x-2\right)<10$

  $\Rightarrow -10+2<x<10+2$

  $\Rightarrow -8<x<12$

Thus, the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|\frac{{\left(x-2\right)}^n}{{10}^n}\right|}$ converges in interval $-8<x<12$ or $x\in \left(-8,12\right)$ .

Since, the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|\frac{{\left(x-2\right)}^n}{{10}^n}\right|}$ is convergent in interval $-8<x<12$ or $x\in \left(-8,12\right)$ so does the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ .

Conclusion:

The interval of convergence of series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is $-8<x<12$ or $x\in \left(-8,12\right)$ .

b.

To determine

To find:The value of x for which the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is absolutely convergent.

Answer to Problem 39E

The series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is absolutely convergent in interval $-8<x<12$ .

Explanation of Solution

Given:

Theseries ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ in part (a).

Concept used:

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

Calculations:

In part (a), we show the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|\frac{{\left(x-2\right)}^n}{{10}^n}\right|}$ converges in interval $-8<x<12$ that implies the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is absolutely convergent in interval $-8<x<12$ .

Conclusion:

The series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is absolutely convergent in interval $-8<x<12$ .

c.

To determine

To find: The value of x for which the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ is conditionally convergent.

Answer to Problem 39E

There is no interval in which the series converge conditionally.

Explanation of Solution

Given:

The series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ 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 ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ and ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|\frac{{\left(x-2\right)}^n}{{10}^n}\right|}$ both converge in the interval $-8<x<12$ .

Since, the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\frac{{\left(x-2\right)}^n}{{10}^n}}$ absolutely convergentin interval, $-8<x<12$ , therefore there is no interval in which the series converge conditionally.

Conclusion:

There is no interval in which the series converge conditionally.