Answer
Converges
Work Step by Step
Consider $u_n=\dfrac{\ln n}{n}$
Here, $u_n$ is positive for all the values of $n$.
Now, $f(n)=\dfrac{\ln n}{n}$
and $f'(n)=\dfrac{1-\ln n}{n^2} \leq 0$ when $n \geq 3$
The negative sign shows that, the sequence $u_n$ is not increasing.
Thus, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty}\dfrac{\ln n}{n}$
Need to apply L-Hospital's rule.
we have $\lim\limits_{n \to \infty}\dfrac{\ln n}{n}=\lim\limits_{n \to \infty}\dfrac{1/n}{1}=0$
Hence, by the Alternating Series Test, the series converges.
