Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{\ln (n+1)}{n+1}$
When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase.
Thus, we have $\dfrac{\ln (n+1)}{n+1} \gt \dfrac{1}{n+1}=\dfrac{1}{n}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent series due to p-series test with $p=1$
Thus the series diverges due to the direct comparison test.
