Chapter 9.5, Problem 25E

To determine

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

Answer to Problem 25E

The series converges conditionally, and the truncation error is less than $0.0022$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\ln n}}$

Calculation:

Let examine whether the series converges absolutely. If it does not, let use other tests in order to determine conditional convergence. The bound for the truncation error can be determined by using the Alternating Series Bound Theorem.

First let examine the series ${\displaystyle \sum _{n=2}^{\infty }\left|a_n\right|}$ . If this series converges, then the series converges absolutely.

  ${\displaystyle \sum _{n=2}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\ln n}}$

In order to determine the convergence or divergence of the series let use the Integral Test.

Based on Integral Test, ${\displaystyle \sum _{n=2}^{\infty }1/(n\ln n)}$ converges if and only if ${\displaystyle \sum _{n=2}^{\infty }1/(x\ln x)dx}$ converges.

Let evaluate the improper integral from the previous step. Let $t=\ln x$ hence $dt=1/xdx$ , and thus

  $\begin{array}{c}{\displaystyle \sum _{n=2}^{\infty }\frac{1}{x\ln x}dx}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{t}dt}\\ ={\underset{k\to \infty }{\lim }}^k\frac{1}{t}dt\\ $ =\underset{k\to \infty }{\lim }{[\ln t]}_2^k$$ =\underset{k\to \infty }{\lim }{[\ln \ln x]}_2^k$=\underset{k\to \infty }{\lim }(\ln \ln k-\ln \ln 2)\\ {\displaystyle \sum _{n=2}^{\infty }\frac{1}{x\ln x}dx}=\infty \end{array}$

because $\ln k$ approaches infinity as $k$ approaches infinity.

Based on the previous step, ${\displaystyle \sum _{n=2}^{\infty }\left|a_n\right|}$ diverges, the original series does not converge absolutely.

Let check whether the series converges conditionally by using the alternating series test.

Since $\ln n$ is positive for every $n\ge 2$ , also $u_n=1/(n\ln n)$ is positive, the first condition is satisfied.

The expression $n\ln n$ is increasing, hence $u_n=1/(n\ln n)$ is decreasing for every $n\ge 2$ , thus also the second condition is satisfied.

Finally, $n\ln n$ grows without bound as $n$ approaches infinity, therefore in the meantime $u_n=1/(n\ln n)$ approaches zero also the third condition is satisfied.

Since $u_n$ satisfies all the three conditions of the the alternating series test, based on the test the series

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

converges conditionally.

The truncation error can be determined by using the Alternating Series Bound Theorem. Based on this, the error after 99 terms is less than $u_{100,}$ thus

  $\begin{array}{c}$ \text{ error }<u_{100}-{(-1)}^{100}\frac{1}{100\cdot \ln 100}$\approx 0.0022\end{array}$

Therefore, the series converges conditionally, and the truncation error is less than $0.0022$ .