Chapter 9.5, Problem 3QQ

To determine

To find: Out of the given four options which of the following describes 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 option $\left(D\right)$ .

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{\ln n}{n}$ is given.

Since ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n}}$ , $f\left(x\right)=\frac{\ln x}{x}$ is a continuous, positive and decreasing function of $x$ on the interval $\left[1,\infty \right)$ .As a result, apply Integral test.

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}f\left(x\right)}={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\ln x}{x}}dx$

Now, consider $\ln x=u$ such that $\frac{1}{x}dx=du$ such that the limit varies from $\infty$ to $0$ . Use the mentioned substitutions and limits to the right side of the obtained equation and simplify.

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\ln x}{x}}dx={\displaystyle \underset{0}{\overset{\infty }{\int }}udu}\\ ={\left[\frac{u^2}{2}\right]}_0^{\infty }\\ =\left[\infty -0\right]\\ =\infty \end{array}$

This implies that the integral does not absolutely converges.

Now, use Alternate series test on $S={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n}\frac{\ln n}{n}=\frac{\ln 2}{2}-\frac{\ln 3}{3}+\frac{\ln 4}{4}-\cdots$ .

• It is evident from the series that $\frac{\ln n}{n}>0$ for $n>1$ .

Differentiate $f\left(x\right)=\frac{\ln x}{x}$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\frac{\ln x}{x}\right)\\ f^{\prime }\left(x\right)=\frac{x\frac{d}{dx}\left(\ln \left(x\right)\right)-\frac{d}{dx}\left(x\right)\ln \left(x\right)}{x^2}\\ =\frac{x\left(\frac{1}{x}\right)-1\left(\ln \left(x\right)\right)}{x^2}\\ =\frac{1-\ln \left(x\right)}{x^2}\end{array}$

This implies that $f^{\prime }\left(x\right)=\frac{1-\ln \left(x\right)}{x^2}<0$

Solve the inequality $\frac{1-\ln \left(x\right)}{x^2}<0$

  $\begin{array}{c}\frac{1-\ln \left(x\right)}{x^2}<0\\ 1-\ln \left(x\right)<0\\ \left(-\ln \left(x\right)\right)\left(-1\right)>\left(-1\right)\left(-1\right)\\ \ln \left(x\right)>1\\ x>e\end{array}$

• This implies that function $f\left(x\right)=\frac{\ln x}{x}$ is decreasing for $x>e$ . Therefore, $\frac{\ln 4}{4}$ is greater than $\frac{\ln 5}{5}$ and so on for the coming terms.

Also, determine $\underset{n\to \infty }{\lim }\frac{\ln n}{n}$ .

  $\underset{n\to \infty }{\lim }\frac{\ln n}{n}=\frac{\infty }{\infty }$

Since the obtained limit is indeterminate form, apply L’Hospital’s Rule to $\underset{n\to \infty }{\lim }\frac{\ln n}{n}$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\ln n}{n}=\underset{n\to \infty }{\lim }\frac{\frac{d}{dn}\left(\ln n\right)}{\frac{d}{dn}\left(n\right)}\\ =\underset{n\to \infty }{\lim }\frac{\frac{1}{n}}{1}\\ =\underset{n\to \infty }{\lim }\frac{1}{n}\\ =0\end{array}$

• Therefore, $\underset{n\to \infty }{\lim }\frac{\ln n}{n}=0$ .

Since all the three conditions of the Alternating series test are satisfied, the given series converges.

It is clearly observable that Integral Series Test shows that the given series does not absolutely converges, but according to the Alternating series test, it converges. As a result, it can be said that the given series converges but conditionally.

Therefore, the correct choice is $\left(D\right)$ .