Chapter 10.5, Problem 23E

To determine

To find: the series converges absolutely converges conditionally or diverges.

Answer to Problem 23E

Conditionally converges

Error $\text{< 0}\text{.0101}$

Explanation of Solution

Given:

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

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}}\frac{1+n}{n^2}=2-\frac{2}{3}+\frac{4}{9}-\frac{5}{16}+\frac{6}{25}-\mathrm{...}\\ \underset{n\to \infty }{\lim }\frac{1+n}{n^2}\end{array}$

Alternating series test:

  $\frac{1+n}{n^2}$ is positive.

The limit of the terms tends to zero.

Each successive term (without the sign) is smaller than the previous term.

So it converges.

To test for absolute convergence

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\frac{1+n}{n^2}={\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n^2}+\frac{n}{n^2}\right)}}\\ ={\displaystyle \sum _{n=1}^{\infty }\left(\frac{1}{n^2}+\frac{1}{n}\right)}\end{array}$

  $\frac{1}{n^2}$ is known to converge and $\frac{1}{n}$ is known to therefore the whole thing divergent.

The original series converges and it doesn’t converge absolutely, so it conditionally converges.

Using the alternative series estimation theorem, therefore

  $\text{error < 0}\text{.0101}$

Hence, the correct option is (b)