Chapter 8.5, Problem 25E

a.

To determine

To identify: Whether the given series are convergent.

Answer to Problem 25E

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-2\right)}^n$ is convergent.

Explanation of Solution

Given information:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n$ is convergent and ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-2\right)}^n$ is given

Concept Used:

Theorem:

For the given power series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(x-a\right)}^n$ there are only three possibilities:

The series converges only when $x=a$

The series converges for all $x$

There is a positive number $R$ such that the series if $\left|x-a\right|<R$ and diverges if $\left|x-a\right|>R$

And the interval of the convergence for the case (iii) is $\left(a-R,a+R\right)$

Calculation:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n$ is convergent.

According to the theorem :

For this question

  $\left|x-a\right|=4$

Which means $R$ is at minimum 4.

  $\begin{array}{l}\left|x-a\right|=4\\ \left|x-a\right|<R\\ 4<R\\ so\\ R>4\end{array}$

Then the minimum interval of convergence be

  $\left(a-4,a+4\right)$

And the radius of convergence is $R\ge 4$

For

  ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-2\right)}^n$

Now,

  $\left|-2\right|=2<4$ which is in the interval of convergence for the minimum $R=4$

So, the series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-2\right)}^n$ also converges

Conclusion:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-2\right)}^n$ is convergent.

b.

To determine

To identify: Whether the given series are convergent.

Answer to Problem 25E

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n$ cannot conclude convergent or divergent.

Explanation of Solution

Given information:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n$ is convergent and ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n$ is given.

Concept Used:

Theorem:

For the given power series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(x-a\right)}^n$ there are only three possibilities:

The series converges only when $x=a$

The series converges for all $x$

There is a positive number $R$ such that the series if $\left|x-a\right|<R$ and diverges if $\left|x-a\right|>R$

And the interval of the convergence for the case (iii) is $\left(a-R,a+R\right)$

Calculation:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n$ is convergent.

According to the theorem :

For this question

  $\left|x-a\right|=4$

Which means $R$ is at minimum 4.

  $\begin{array}{l}\left|x-a\right|=4\\ \left|x-a\right|<R\\ 4<R\\ so\\ R>4\end{array}$

Then the minimum interval of convergence be

  $\left(a-4,a+4\right)$

And the radius of convergence is $R\ge 4$

For

  ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n$

Now,

  $\left|-4\right|=4$ which is at the extreme of the interval of convergence.

So

Consider the following cases:

Case $1$ :

  $c_n=\frac{1}{n{\left(-4\right)}^n}$

Then the main series become

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n{\left(-4\right)}^n}\cdot 4^n}\\ ={\displaystyle \sum _{n=0}^{n=\infty }\frac{{\left(-1\right)}^n}{n}}\end{array}$

This converges by the alternating series test.

The given series becomes

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n{\left(-4\right)}^n}\cdot {(-4)}^n}\\ ={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n}}\end{array}$

This is harmonic series so it is divergent.

Case $1$ :

  $c_n=\frac{1}{n^2{\left(-4\right)}^n}$

Then the main series become

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(4\right)}^n={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n^2{\left(-4\right)}^n}\cdot 4^n}\\ ={\displaystyle \sum _{n=0}^{n=\infty }\frac{{\left(-1\right)}^n}{n^2}}\end{array}$

This converges by the alternating series test.

The given series becomes

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n^2{\left(-4\right)}^n}\cdot {(-4)}^n}\\ ={\displaystyle \sum _{n=0}^{n=\infty }\frac{1}{n^2}}\end{array}$

This is convergent series .

So, the series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n$ cannot conclude to be convergent or divergent.

Conclusion:

The series ${\displaystyle \sum _{n=0}^{n=\infty }{c}_n}{\left(-4\right)}^n$ cannot conclude convergent or divergent.