convergence of $\sum\frac{\sin(\sqrt{n})}{n^a}$ For $a>0$, define $x_n=\dfrac{\sin(\sqrt{n})}{n^a}$.
Prove that $\sum x_n$ converges if and only if $a>\dfrac1 2$.
I proved the convergence when $a>\dfrac1 2$, but I can't derive the other direction. Could someone give me some help? Thanks a lot.

A straightforward way to show that this series diverges if $a \leqslant 1/2$ is to show that the Cauchy criterion is not satisfied.
If $\sqrt{n} \in [c_1,c_2] = [\pi/6 + 2k\pi, 5\pi/6 + 2k\pi]$ then $\sin \sqrt{n} \geqslant 1/2$ and, with $a \leqslant 1/2$, we have
$$\begin{align} \left|\sum_{c_1 \leqslant \sqrt{n} \leqslant c_2}\frac{\sin \sqrt{n}}{n^a}\right| &= \left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| \\  &\geqslant \frac{1}{2} \frac{c_2^2 - c_1^2}{c_2^{2a}} \\ &\geqslant \frac{1}{2}\frac{c_2^2 - c_1^2}{c_2}  \\ &= \frac{1}{2}\frac{\frac{8 \pi^2k}{3} + \frac{2\pi^2}{3}}{\frac{5\pi}{6 }+ 2 k \pi} \end{align}$$
The expression on the right-hand side converges to $2 \pi/3$ as $k \to \infty$. For any $N \in \mathbb{N}$ there exists sufficiently large $k \in \mathbb{N}$ such that $c_2^2 > c_1^2 > N$
and
$$\left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| > \frac{\pi}{3}.$$
Thus the series fails to satisfy the Cauchy criterion and is divergent if $a \leqslant 1/2.$
