Does the series $\sum\frac{\ln n}{n^{2}}$ Converge? Does the series
$$\sum_{n=2}^{\infty}\frac{\ln n}{n^2}$$
converge?
I'm searching for a solution that does not use the Integral test, Stirling, L’Hôpital or functions theorems.
I tried ratio test, and also comparing it to another series.
Thank you very much.
 A: 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 $q=1/2$.
A: 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 $-\zeta^{'}(2).$
A: You can show that $\ln n\leq \sqrt n$ if $n$ is large enough. Now you can readily deduce that the series converges.
