How to prove that $\sum_{n=1}^{\infty} \frac{(\log (n))^2}{n^2}$ converges? $$\sum_{n=1}^{\infty} \frac{(\log (n))^2}{n^2}$$
I know that this series converges (proof by Answer Sheet). However I need to prove it using comparison, integration, ratio or other tests. 

The integration test doesn't seem to help.
The ratio test seemed to shed light except that it requires further proofs that $\frac{log(n+1)}{log(n)} < 1$ etc which makes me think this is not the best approach.
I considered the fact that $\log(n) < \sqrt{n}$ but this just shows that it is less than a divergent series which doesn't help.
Suggestions?
 A: Have you tried Cauchy's condensation test?
Remark : This test is usually useful to get rid of logarithms when trying to check if a series converges or diverges. 
Hope that helps,
A: This is a Bertrand's series. You can easily prove convergence using asymptotic analysis.
Indeed, we have $\log^2n=_\infty o(n^{1/2})$, whence 
$$\frac{\log^2n}{n^2}=\frac{o(n^{1/2})}{n^2}=o\biggl(\frac1{n^{3/2}}\biggr).$$
As both are series with positive terms and the latter converges, the former does too.
A: There is of course the homely old Integral Test:
$$ \ \int_1^{\infty} \ \ \frac{(\log x)^2}{x^2} \ \ dx  \ \ = \ \ \left[ - \ \frac{(\log x)^2 \ +  \ 2 \log x \ + \ 2 \ }{x} \right]_1^{\infty}  \ \ = \ \ 2 \ \ . $$
I'm not sure what the remark about the "integration test doesn't seem to help" is intended to mean: one must just be a bit patient with integration-by-parts and l'Hopital or some other limit technique.  
In fact,  one finds that the result can be generalized to
$$ \ \int_1^{\infty}  \ \frac{(\log x)^p}{x^q} \ \ dx  \ \ $$ 
convergent for integers $ \ p \ \ge \ 1 \ $ and $ \ q \ \ge \ 2 \ $ .  We have   $$ \ \int_1^{\infty} \ \ \frac{(\log x)^p}{x^2 } \ \ dx  \ \ = \ \ p! \ \ , $$
[EDIT: This last result can be demonstrated by connecting the reduction formula,
$$ \ \int  \ \ \frac{(\log x)^p}{x^2 } \ \ dx  \ \ = \ \ -\frac{(\log x)^p}{x} \  \ + \ \ p \ \int  \ \ \frac{(\log x)^{p-1}}{x^2 } \ \ dx \ \ , $$
with our earlier expression for $ \ p \ = \ 2 \ $ . ]
and  $$   \frac{(\log x)^p}{x^2 } \ \ge \ \frac{(\log x)^p}{x^q }    $$
for $ \   q \ > \ 2 \ $ and $ \ x \ \ge \ 1 \ $ .  This establishes our convergence proposition for the improper integrals (by integral comparison), so  
$$ \sum_{n=1}^{\infty} \frac{(\log n)^p}{n^q} $$
converges for integers $ \ p \ \ge \ 0 \ $ and $ \ q \ \ge \ 2 \ $ .
A: By comparison test we have $$\frac{n^{3/2}(\log (n))^2}{n^2}= \frac{(\log (n))^2}{n^{1/2}} =\left(\frac{\log (n)}{n^{1/4}}\right)^2=\left(4\frac{\log (n^{1/4})}{n^{1/4}}\right)^2\to 0$$ that is for n large enough we have
$$\frac{(\log (n))^2}{n^2}<\frac{1}{n^{3/2}}$$
the convergence follow by Riemann series. see more general here  On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$
