Chapter 10.5, Problem 22E

To determine

To find: whether the series converges or diverges.

Answer to Problem 22E

Since, the integral diverges, hence from the statement of Integral test series also diverges.

Explanation of Solution

Given information:

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

Calculation:

The 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.

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

Here, the Integral Test applies because

  $f(x)=\frac{1}{x}-\frac{1}{x^2}$

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

We have,

  $\begin{array}{cc}{{\displaystyle \int }}_1^{\infty }\left(\frac{1}{x}-\frac{1}{x^2}\right)dx& =\underset{k\to \infty }{\lim }{{\displaystyle \int }}_1^k\left(\frac{1}{x}-\frac{1}{x^2}\right)dx\\ & =\underset{k\to \infty }{\lim }{\left[\ln x+\frac{1}{x}\right]}_1^k\\ & =\underset{k\to \infty }{\lim }\left[\ln k-\ln 1+\frac{1}{k}-1\right]\\ & =\infty \end{array}$

Since, the integral diverges, hence from the statement of Integral test series also diverges.