Show the series $\sum \frac{\log n}{n^2}$ converges by comparison I know that 
$$
\sum \frac{\log n}{n^2}
$$
Converges by the integral test
$$
\int_1^{\infty}\frac{\log x}{x^2}dx=-\frac{\log x}{x}+\int_1^{\infty}\frac{1}{x^2}dx=c-\lim_{x\rightarrow}\frac{\log x}{x}=c
$$
But I was curious as to whether there was a clever way to show convergence by comparison. 
 A: For large $n$, $\log n < n^{\frac{1}{2}}$. And, 
$$
\sum\frac{n^{\frac{1}{2}}}{n^{2}}=\sum\frac{1}{n^{\frac{3}{2}}}<\infty,
$$
by the $p$-test since $\frac{3}{2}>1$.
Therefore, the original series converges.  
A: $$\frac{\log n}{n^2}\le\frac{n^{1/2}}{n^2}=\frac1{n^{3/2}}$$
Because for any $\;\epsilon>0\;$, we have that for all but a finite number of values of 
$$\;n\in\Bbb N\;,\;\;\;\log n<n^\epsilon$$
A: An unusual proof may go through the following lines: since for any $n\geq 1$ we have
$$ \log(n) \leq H_n-\gamma-\frac{1}{2(n+1)} \tag{1}$$
it follows that:
$$ \sum_{n\geq 1}\frac{\log n}{n^2}\color{red}{\leq} \sum_{n\geq 1}\frac{H_n}{n^2}-\gamma\zeta(2)-\sum_{n\geq 1}\frac{1}{2n^2(n+1)}=\color{red}{\frac{6+24\,\zeta(3)-\pi^2(1+2\gamma)}{12}}\tag{2} $$
A: We can also use the Cauchy (Schlömilch) condensation test and then the comparison test.
\begin{equation}
\sum\limits_{k=1}^{\infty} \frac{\mathrm{e}^{k}\mathrm{ln}(\mathrm{e}^{k})}{\mathrm{e}^{2k}}
= \sum\limits_{k=1}^{\infty} \frac{k}{\mathrm{e}^{k}}
\end{equation}
We then compare this series with the convergent series $$\sum\limits_{k=1}^{\infty} \frac{k}{k^{3}}$$ and
thus our original series converges.
