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.

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.

For $k\in \Bbb N$ let $n_k=\lfloor (k+\frac12)^2\pi^2\rfloor$. Then $\sin^2 \sqrt n_k\approx 1$ and also $\sin^2 \sqrt {n_k+d}\approx 1$ for $0<d<\sqrt{n_d}$. This gives us $\approx k\pi$ summands of size $\approx \frac1{k^2\pi^2}$, i.e., a contribution of $\approx \frac1{k\pi}$. This allows us to compare with the divergent harmonic series.

While the "$\approx$" used in this argument should be made more explicit for a formal proof, we can be quite generous at this ...