Answer
Diverges, by Th. 9.12.
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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Observe that for $n\geq 3,\quad \ln n > \ln e=1$
$\displaystyle \frac{\ln n}{n+1} > \frac{1}{n+1} > 0$.
$(b_{n} > a_{n} > 0)$.
$\displaystyle \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{1}{n+1} = \displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$
$=(1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)-1$
$=(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n})-1$,
diverges (harmonic, divergent p-series)
So, case 2 of Th.9.12 applies:
.
Since $\displaystyle \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{1}{n+1}$ diverges, then
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{\ln n}{n+1}$ also diverges.