Chapter 10.5, Problem 3QQ

To determine

The behaviour of the series ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{\ln n}{n}$

Answer to Problem 3QQ

The correct option is D

Explanation of Solution

Given information:

The given series is,

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

Formula used:

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

Calculation:

Using integral test series

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n}}\\ {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\ln x}{x}}=\underset{k\to \infty }{\lim }{\left[{\left(\ln x\right)}^2\right]}_1^4=\infty \end{array}$

So,

  ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{\ln n}{n}$ does not absolutely converge.

By using Alternate Series Test, $S={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{\ln n}{n}=\frac{\ln 2}{2}-\frac{\ln 3}{3}+\frac{\ln 4}{4}-\mathrm{......}$

Since $\frac{\ln n}{n}>0$ for $n>\mathrm{1........}\left(1\right)$

Take $f\left(x\right)=\frac{\ln x}{x}$

Now differentiating $f\left(x\right)$ with respect to x, $f\text{'}\left(x\right)=\frac{1-\ln x}{x^2}$

  $f\text{'}\left(x\right)=\frac{1-\ln x}{x^2}<0,x>e$

That is $f\left(x\right)$ decreasing for $x>e$ .

Therefore, $\frac{\ln 4}{4}$ is greater than from $\frac{\ln 5}{5}$ and so on....

Now $\underset{n\to \infty }{\lim }\frac{\ln n}{n}=\mathrm{0.........}\left(3\right)$

Therefore, from equation (1), (2) and (3), the series ${{\displaystyle \sum _{n=1}^{\infty }\left(-1\right)}}^n\frac{\ln n}{n}$

Converges conditionally

Therefore, answer is D

Conclusion:

From equation (1), (2) and (3), the series ${{\displaystyle \sum _{n=1}^{\infty }\left(-1\right)}}^n\frac{\ln n}{n}$ Converges conditionally