Does $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2+2}$ converge? I'm trying to find out whether
$$\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2+2}$$
is convergent or divergent?

The series can be estimated termwise as follows:
\begin{eqnarray}
\frac{\ln(n)}{n^2 + 2} & \leq & \frac{\ln(n)}{n^2} = \ln(n) e^{-2\ln(n)} = 2 \cdot \frac{1}{2} \ln(n) e^{-2\ln(n)} \leq 2 \cdot \bigg[ \sum_{k=0}^\infty \frac{1}{k!}\bigg(\frac{1}{2} \ln(n)\bigg)^k \bigg] e^{-2\ln(n)} \\
& = & 2 \cdot e^{\frac{1}{2}\ln(n)} e^{-2\ln(n)} = 2 \cdot e^{-\frac{3}{2} \ln(n)} = 2 \cdot n^{-3/2} \ .
\end{eqnarray}
Now the series converges because it is non-negative and has a converging majorant.
