Chapter 9.5, Problem 22E

To determine

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

Answer to Problem 22E

The series diverges.

Explanation of Solution

Given:

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

Formula used:

Integral test Let $\left\{a_n\right\}$ be a sequence of positive terms. Suppose that $a_n=f(n)$ , where $f$ is a continuous positive, decreasing function of $x$ for all $x\ge N$ ( $N$ 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 converge or both diverge.

Calculation:

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

The integral test applies because,

  $f(x)=\frac{1}{x}-\frac{1}{x^2}$

It is a continuous positive, decreasing function of $x$ for $x\ge 1$ .

  $\begin{array}{l}{\displaystyle {\int }_N^{\infty }\frac{1}{x}-\frac{1}{x^2}dx}=\underset{n\to \infty }{\lim }{\displaystyle {\int }_N^{\infty }\frac{1}{x}-\frac{1}{x^2}dx}\\ {\displaystyle {\int }_N^{\infty }\frac{1}{x}-\frac{1}{x^2}dx}=\infty \end{array}$

The integral diverges, from the statement of integral test. Thus series diverges.