<s>[INST] Does the series $\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$ converge? 
Does the following series converge? 
  $$\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$$

It shouldn't, but I have no idea how to prove it. I was wondering about Integral Criterion, but the assumptions are not satisfied. Or perhaps Dirichlet test would help, but then it should be shown that $\sum_{k=1}^n(\sin^2(\sqrt{k}))$ is bounded.
 [/INST] Observe that $\sin^2(\sqrt{n}) \geq 1/4$ iff $|\sin(\sqrt{n})| \geq 1/2$ iff 
$$\frac{\pi}{6} + k\pi \leq \sqrt{n} \leq \frac{5\pi}{6} + k\pi$$
for some nonnegative integer $k$. This chain of inequalities is equivalent to
$$\left(\frac{\pi}{6} + k\pi\right)^2 \leq n \leq \left(\frac{5\pi}{6} + k\pi\right)^2$$
For a fixed $k$, the number of values of $n$ which satisfy the above is approximately
$$\left(\frac{5\pi}{6} + k\pi\right)^2 - \left(\frac{\pi}{6} + k\pi\right)^2
= \frac{2\pi^2}{3} + \frac{4\pi^2}{3}k > 6+13k$$
Therefore, 
$$\sum_{n=1}^{\infty}\frac{\sin^2(\sqrt{n})}{n} > \sum_{k=1}^{\infty}\frac{6+13k}{4}\frac{1}{\left(\frac{5\pi}{6} + k\pi\right)^2}$$
which diverges by limit comparison with $\sum\frac{1}{k}$.
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