Does the series $\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$ converge?

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 $\sum\frac{1}{k}$.

Denote $S_N=\sum_{n=1}^N\frac{\sin^2(\sqrt{n})}{n}$.

By the Euler-MacLaurin formula we have

$$ S_N\sim_{\infty}\int_1^N dx\frac{\sin^2(\sqrt{x})}{x}+\mathcal{O}(1) $$

You can show this by observing that the derivates of $\frac{\sin^2(\sqrt{x})}{x}$ are $\mathcal{O}\left(\frac{1}{x^{1+m/2}}\right)$, where $m$ is the order of the derivative.

Performig a change of variables $x=y^2$ we get $$S_N\sim_{\infty}2\int_1^N dx\frac{\sin^2(y)}{y}+\mathcal{O}(1)\sim\log(N)+\mathcal{O}(1) $$

which shows that the sum is unbounded. The last asymptotic identity can be proved by using $\sin(x)^2=\frac{1}{2}(1-\cos(2x))$ combined with an integration by part.