Convergence of $\frac{\ln(n)}{n^2}$ During a Calc exam, I needed to state whether $$\sum_{n=1}^{∞} \frac{\ln n}{n^2}$$ converged or diverged. I was going to try a strict comparison test (with $\sum_{n=3}^{∞} \frac{\ln n}{n^2}$ to $\sum_{n=3}^{∞} \frac{1}{n^2}$) but I realized this wouldn't work because $\frac{1}{n^2}$ converges but is less than $\frac{\ln n}{n^2}$ (and it needs to be greater in this case). So I decided to try using a limit comparison with $\frac{1}{n^\frac{3}{2}}$. So I got
$$\lim_{n\to ∞}{\frac{\frac{\ln n}{n^2}}{\frac{1}{n^\frac{3}{2}}}} = \lim_{n\to ∞}{{\frac{\ln n}{n^2}}{\frac{n^\frac{3}{2}}{1}}} = \lim_{n\to ∞}{\frac{\ln n}{n^\frac{1}{2}}}= \frac{∞}{∞}$$ and using l'Hopital's Rule ...
$$\lim_{x\to ∞}{\frac{\frac{1}{x}}{\frac{1}{2x^\frac{1}{2}}}}=\lim_{x\to ∞}{\frac{2x^\frac{1}{2}}{x}} = \lim_{x\to ∞}{\frac{2}{x^\frac{1}{2}}} = 0.$$
Therefore the series must converge using the limit comparison test. 
So was this the correct way to solve this?

The series $\displaystyle \sum_{n \geq 1} \frac{1}{n^{\alpha}}$ converges if and only if $\alpha > 1$. 
Note that :
$$ \frac{\ln(n)}{n^{2}} n^{\alpha} = n^{\alpha-2} \ln(n). $$
For any $\alpha$ such that $\alpha > 1$ and $\alpha-2 < 0$ ($\displaystyle \alpha = \frac{3}{2}$ works!), we have :
$$ \lim \limits_{n \to +\infty} n^{\alpha-2}\ln(n) = 0 $$
which ensures that :
$$ \frac{\ln(n)}{n^{2}} = \mathop{o} \limits_{n \to +\infty} \Big( \frac{1}{n^{\alpha}} \Big). $$
Since the series $\displaystyle \sum_{n \geq 1} \frac{1}{n^{\alpha}}$ converges (because $\alpha > 1$), by comparison, the series $\displaystyle \sum_{n \geq 1} \frac{\ln(n)}{n^2}$ converges too.
