Chapter 9.5, Problem 1QQ

a.

To determine

To find: To find whether the given series converges or diverges.

Answer to Problem 1QQ

The series converges.

Explanation of Solution

Given:

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

Formula used:

  $a=f(n)$where $f$ is a continuous, positive, decreasing function of x for all $x\ge N$ (N is a positive integer).

Then the series ${\displaystyle \sum _{n=N}^{\infty }{a}_n}$ and the integral ${\displaystyle {\int }_N^{\infty }f(x)dx}$ either both converges or both diverges.

Calculation:

Given series is ${\displaystyle \sum _{n=N}^{\infty }\frac{2}{n^2+1}}$

Using Limit comparing test,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\frac{2}{n^2+1}}{\frac{1}{n^2}}=\underset{x\to \infty }{\lim }\frac{2x^2}{n^2+1}\\ \underset{x\to \infty }{\lim }\frac{\frac{2}{n^2+1}}{\frac{1}{n^2}}=2\end{array}$

The integral test applies because ${\displaystyle {\int }_N^{\infty }\frac{1}{x}dx}$

By p-series test, ${\displaystyle {\int }_N^{\infty }\frac{1}{x^p}dx}$ converges if $p>1$ , and diverges if $p\le 1$ .

It is of the form of ${\displaystyle {\int }_N^{\infty }\frac{1}{x^p}dx}$ where $p=\frac{4}{3}$ .

This series converges since $p>1$ .

The limit is positive, ${\displaystyle \sum _{n=N}^{\infty }\frac{2}{n^2+1}}$ also converges.

b.

To determine

To find: To find whether the given series converges or diverges.

Answer to Problem 1QQ

The series converges.

Explanation of Solution

Given:

The given series ${\displaystyle \sum _{n=N}^{\infty }\frac{2^n-1}{3^n+1}}$ .

Formula used:

  $a=f(n)$where $f$ is a continuous, positive, decreasing function of x for all $x\ge N$ (N is a positive integer).

Then the series ${\displaystyle \sum _{n=N}^{\infty }{a}_n}$ and the integral ${\displaystyle {\int }_N^{\infty }f(x)dx}$ either both converges or both diverges.

Calculation:

Given series is ${\displaystyle \sum _{n=N}^{\infty }\frac{2^n-1}{3^n+1}}$ .

Using Limit comparing test,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\frac{2^n-1}{3^n+1}}{\frac{2^n}{3^n}}=\underset{x\to \infty }{\lim }\frac{\frac{2^n-1}{2^n}}{\frac{3^n+1}{3^n}}\\ \underset{x\to \infty }{\lim }\frac{\frac{2^n-1}{3^n+1}}{\frac{2^n}{3^n}}=1\end{array}$

If the sequence of partial sum has a limit as $n\to \infty$ , series convergence otherwise the sequence is divergence.

  $S={\displaystyle \sum _{n=N}^{\infty }\frac{2^n}{3^n}}$

This is a geometric series whose initial term is zero and common ratio is $\frac{1}{2}$ .

  $\begin{array}{l}S={\displaystyle \sum _{n=N}^{\infty }\frac{2^n}{3^n}}=\frac{\frac{2}{3}}{1-\frac{2}{3}}\\ S=2\end{array}$

The limit is positive, ${\displaystyle \sum _{n=N}^{\infty }\frac{2}{n^2+1}}$ also converges.

c.

To determine

To find whether the given series converges or diverges.

Answer to Problem 1QQ

The series converges.

Explanation of Solution

Given:

The given series ${\displaystyle \sum _{n=N}^{\infty }\frac{\sqrt[4]{n}}{n}}$ .

Formula used:

  $a=f(n)$where $f$ is a continuous, positive, decreasing function of x for all $x\ge N$ (N is a positive integer).

Then the series ${\displaystyle \sum _{n=N}^{\infty }{a}_n}$ and the integral ${\displaystyle {\int }_N^{\infty }f(x)dx}$ either both converges or both diverges.

Calculation:

Given series is ${\displaystyle \sum _{n=N}^{\infty }\frac{\sqrt[4]{n}}{n}}$ .

  ${\displaystyle \sum _{n=N}^{\infty }\frac{\sqrt[4]{n}}{n}}$

By p-series test, ${\displaystyle {\int }_N^{\infty }\frac{1}{x^p}dx}$ converges if $p>1$ , and diverges if $p\le 1$ .

It is of the form of ${\displaystyle {\int }_N^{\infty }\frac{1}{x^p}dx}$ where $p=\frac{3}{4}$ .

This series converges since $p=1$ .

The limit is positive, ${\displaystyle {\int }_N^{\infty }\frac{\sqrt[4]{n}}{n}dx}$ also divergence.