Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges 
Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges

I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.
Any hints? 
Thank you! 
 A: Hint:
$$\sum_{k=1}^{\infty} \frac{\text{ln}\ k}{k^2} < \int_{1}^{\infty}\frac{\text{ln}\ x}{x^2} dx = 1 $$
A: *

*Try looking at the maximum value of the function $\displaystyle f(x) = \frac{\ln(x)}{\sqrt{x}}$ on $(1,\infty)$

*$\displaystyle f'(x)= \frac{2-\ln(x)}{2x^{3/2}}$

*$f'(x) <0$ for all $x>e^{2}$. Hence $f(x) \leq f(e^{2})$ for all $x\geq 1$. Hence $\displaystyle \frac{\ln(x)}{\sqrt{x}} \leq \frac{2}{e}$.

*Now $\displaystyle \sum_{k=1}^{\infty} \frac{\ln{(k)}}{k^{2}} = \sum\limits_{k=1}^{\infty} \frac{\ln(k)}{\sqrt{k}} \times \frac{1}{k\sqrt{k}} \leq \frac{2}{e} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} <\infty$
A: Since
$
\mathop {\lim }\limits_{n \to \infty } \frac{{\ln k}}
{{\sqrt k }} = 0
$ you have that


*

*There exists $k_0$ such that for each $k>k_0$  it is $
\frac{{\ln k}}
{{\sqrt k }} < 1
$

*This means that for each $k>k_0$ it is 
$
\frac{{\ln k}}
{{k^2 }} < k^{ - \frac{3}
{2}} 
$

*Since the series with general term $
k^{ - \frac{3}
{2}} 
$ is convergent your series is also convergent by comparison test.

A: Equals the constant in http://oeis.org/A073002 , which gives further links.
