Chapter 9.5, Problem 23E

To determine

To calculate: The series converges absolutely, converges conditionally, or diverges.

Answer to Problem 23E

The series is conditionally converges, $\text{error }<0.0101$ .

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{1+n}{n^2}}=2-\frac{3}{4}+\frac{4}{9}-\frac{5}{16}+\frac{6}{25}-\dots \\ \underset{n\to \infty }{\lim }\frac{1+n}{n^2}=0\end{array}$

Alternating series test

  $\frac{1+n}{n^2}$ is positive.

The limit of the terms approaches zero.

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

So it converges.

To test for absolute convergence, drop the (-1) part, as the rest is positive. Separate the fraction

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

  $\frac{1}{n^2}$ is known to converge and $\frac{1}{n}$ is known to diverge Just having one part diverge makes the whole thing divergent.

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

Using the alternating series estimation theorem, the truncation error after 99 terms is less than the value of the 100th term.

  $\begin{array}{c}\text{error }<\left|\frac{1+100}{{100}^2}\right|\\ \text{error }<0.0101\end{array}$

Therefore, the series is conditionally converges, $\text{error }<0.0101$ .