Does the series $\sum\frac{\ln n}{n^{2}}$ Converge?

You can show that $\ln n\leq \sqrt n$ if $n$ is large enough. Now you can readily deduce that the series converges.

You can use the condensation test for series whose terms are positive and weakly decreasing.

You replace the sum $$\sum_n a_n$$ by $$\sum_k 2^k a_{2^k}$$ and check that it converges.

Edited to add details:

$$\sum_{n=1}^{\infty}\frac{\ln n}{n^2} \longrightarrow \sum_{k=0}^{\infty}2^k \frac{\ln 2^k}{2^{2k}}=\sum_{k=0}^{\infty}\frac{k \ln 2}{2^k}$$

The last sum converges by the ratio test, or by identifying it as the derivative of a geometric series evaluated at $q=1/2$.

Here is a different style of answer, which may be useful to some.

$\zeta(s)=\sum_{n=1}^\infty n^{-s}$, the Riemann zeta function, is analytic on $\Re(s)>1$. This is because on any half plane $\Re(s)>1+\epsilon$, it is a uniformely convergent series of analytic functions. Its derivative, $$\zeta^{'}(s)=-\sum_{n=1}^\infty \log n n^{-s}$$ then also converges on all of $\Re(s)>1$. Now, notice that your sum is $-\zeta^{'}(2).$