Series convergence ques I was just solving another question regarding this topic and just wanna check if I am doing it right.

So the question is whether the given series converges or not, and justify.

$$\sum_{n = 1}^{\infty} \frac{(\ln{n})^2}{ n^2}$$

I have tried using ratio and comparison test, but that leads me nowhere.
For comparison test I compared the term with $\frac{1}{n^2}$. Am I doing that wrong?

Thank you so much :)
Recall that $f(s) = \sum_{n=1}^{\infty} n^{-s}$ converges when $s&gt;1$.

Then we observe that $f''(s)= \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^s}$, with the same region of convergence.

Hence $f''(2) = \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^2}$.
$\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}&lt; \sum_{n=1}^{\infty}\frac{ln(n) \sqrt{n}}{n^2}&lt;\sum_{n=1}^{\infty}\frac{ln(n)}{n^{\frac{3}{2}}}&lt;\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$ 

This is a p-series with p>1 so it converges. Thus $\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}$ converges as well.