Chapter 10.5, Problem 73E

(a)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{2^n}}$ converges or diverges.

Answer to Problem 73E

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{2^n}}$ converges.

Explanation of Solution

Given information:

To test the convergence of a series we have the nth root Test as stated below:

Let ${\displaystyle \sum a_n}$ be a series with $a_n\ge 0$ for $n\ge N$ and suppose that

  $\underset{n\to \infty }{\lim }\sqrt[n]{a_n}=L$ , then

The series converges if $L<1,$

The series diverges if $L<1,$ or L is infinite

And the test is inconclusive if L=1

Formula used:

Ratio test is used.

Calculation:

The given series is

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{2^n}}$

Then,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\sqrt[n]{a_n}=\underset{n\to \infty }{\lim }\sqrt[n]{\frac{n^2}{2^n}}\\ =\underset{n\to \infty }{\lim }\frac{{\left(\sqrt[n]{n}\right)}^2}{2}\\ \underset{n\to \infty }{\lim }\sqrt[n]{a_n}=\frac{1}{2}\end{array}$

The series converges

Conclusion:

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{2^n}}$ converges.

(b)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{n}{2n-1}\right)}^n}$ converges or diverges.

Answer to Problem 73E

The series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{n}{2n-1}\right)}^n}$ converges

Explanation of Solution

Given information:

To test the convergence of a series we have the nth root Test as stated below:

Let ${\displaystyle \sum a_n}$ be a series with $a_n\ge 0$ for $n\ge N$ and suppose that

  $\underset{n\to \infty }{\lim }\sqrt[n]{a_n}=L$ , then

The series converges if $L<1,$

The series diverges if $L<1,$ or L is infinite

And the test is inconclusive if L=1

Formula used:

Ratio test is used.

Calculation:

The given series is

  ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{n}{2n-1}\right)}^n}$

Then,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\sqrt[n]{a_n}=\underset{n\to \infty }{\lim }\sqrt[n]{{\left(\frac{n}{2n-1}\right)}^n}\\ =\underset{n\to \infty }{\lim }\frac{n}{2n-1}\\ \underset{n\to \infty }{\lim }\sqrt[n]{a_n}=\frac{1}{2}\end{array}$

The series converges.

Conclusion:

The series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{n}{2n-1}\right)}^n}$ converges

(c)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges or diverges.

Answer to Problem 73E

The series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges

Explanation of Solution

Given information:

To test the convergence of a series we have the nth root Test as stated below:

Let ${\displaystyle \sum a_n}$ be a series with $a_n\ge 0$ for $n\ge N$ and suppose that

  $\underset{n\to \infty }{\lim }\sqrt[n]{a_n}=L$ , then

The series converges if $L<1,$

The series diverges if $L<1,$ or L is infinite

And the test is inconclusive if L=1

Formula used:

Ratio test is used.

Calculation:

The given series is

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n},\text{where}{a}_n=\{\begin{array}{cc}n/2^n& n\text{is odd}\\ 1/2^n& n\text{is even}\end{array}$

Then,

  $\begin{array}{l}\underset{n\to \infty ,nodd}{\lim }\sqrt[n]{a_n}=\underset{n\to \infty ,nodd}{\lim }\sqrt[n]{\frac{n}{2^n}}\\ =\underset{n\to \infty ,nodd}{\lim }\frac{\sqrt[n]{n}}{2}\\ \underset{n\to \infty ,nodd}{\lim }\sqrt[n]{a_n}=\frac{1}{2}\end{array}$

And

  $\begin{array}{l}\underset{n\to \infty ,neven}{\lim }\sqrt[n]{a_n}=\underset{n\to \infty ,neven}{\lim }\sqrt[n]{\frac{1}{2^n}}\\ \underset{n\to \infty ,neven}{\lim }\sqrt[n]{a_n}=\frac{1}{2}\end{array}$

So the series converges.

Conclusion:

The series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges