Find if the series $\sum_1^\infty (-1)^n\frac{\ln (n)}{n}$ converges or diverges

Question

Find if the series $\sum_1^\infty (-1)^n\frac{\ln (n)}{n}$ converges or diverges.

My attempt: This is an alternating series problem in the form $(-1)^{n-1}b_n$

Let $b_n=\frac{\ln (n)}{n}$

Considering related function $f(x)=\frac{\ln (x)}{x}$;$f'(x)= \frac{1-\ln x}{x^2} $

$f'(x)<0$ when $\ln x>1\rightarrow x>e $. Therefore, $b_n$ is decreasing function when $n>e$

$\lim \limits_{n \to\infty}b_n=\lim \limits_{n \to\infty}\frac{\ln(n)}{n}=\frac{1/n}{1}=\frac{1}{n}\rightarrow0$ as $n\rightarrow \infty$

Therefore series converges by alternating series test. Am I on the right track? I was not sure if I can apply thr alternating series test here as $b_n$ is not decreasing for all values of n. Any help is much appreciated.

Answer 1

Note that

$$\frac{\ln (x)}{x^a}\to 0$$

for all $a>0$.

It is a standard limit which can be easily proved by l’Hopital.

Thus you are done and the series converges.

Answer 2

It is correct. What you did proves that the series $\displaystyle\sum_{n=3}^\infty(-1)^n\frac{\ln n}n$ converges. It follows from this fact that your series converges too.

Answer 3

You are exactly right, alternating series test on a series that goes to 0. You are on the right track. The limit as n goes to infinity goes to 0 and the fact that $b_n$ is decreasing is all you need.