Convergence of $\sum _{k=1}^\infty \sin \left(\sqrt{k}\right)/k$ This was asked at an oral examination.

Does the series  $\displaystyle \sum _{k\geq1}\frac{\sin\left(\sqrt{k}\right)}{k}$ converge ?

After playing with Mathematica, it's very likely it converges, but slowly (sort of oscillating).
To actually prove convergence, summation by part is useless since  $\displaystyle \sum _{k\geq1}\sin\left(\sqrt{k}\right)$ diverges.
Any suggestion is appreciated.
 A: $\sum_{k=1}^{n}\dfrac{\sin(\sqrt{k})}{k}$ plotted against $C-\dfrac{2\cos(\sqrt{k-\sqrt{\pi}})}{\sqrt{k+\sqrt{\pi}}}$, where $C$ is calculated in Mathematica numerically with fact that $\sum_{k=1}^{\infty}\dfrac{\sin(\sqrt{k})}{k}=\sum_{k=1}^{\infty}\dfrac{e^{i\sqrt{k}}}{k}$:

c = Quiet[N[Im[Sum[E^(I Sqrt[k])/k, {k, 1, Infinity}]]]];
Show[Plot[Sum[Sin[Sqrt[k]]/k, {k, 1, n}], {n, 0, 100}, 
GridLines -> {{}, {c}}, GridLinesStyle -> Darker[Green], 
PlotStyle -> {Darker[Green], Thick}], 
Plot[c - 2 Cos[Sqrt[k - Sqrt[Pi]]]/Sqrt[k + Sqrt[Pi]], {k, 0, 100}, 
PlotStyle -> {Darker[Blue], Thin}, PlotRange -> All]]

and for larger $k$:

Quiet[N[Im[Sum[E^(I Sqrt[k])/k, {k, 1, Infinity}]]]]

then gives $C\approx1.7156717726570607\dots$.
A: The convergence of $\int_1^\infty\frac{\sin\sqrt t}tdt$ can be deduced by a substitution $s=\sqrt t$ and the convergence of $\int_1^\infty\frac{\sin s}sds$. Define 
$$a_k:=\int_k^{k+1}\frac{\sin\sqrt t}{t}\mathrm dt-\frac{\sin\sqrt k}k$$
and $g(x):=\frac{\sin(\sqrt x)}x$. Since $$g'(x)=\frac{\cos(\sqrt x)}{x^{3/2}}-\frac{\sin(\sqrt x)}{x^2}$$
and by the mean value theorem, $a_k=g'(x_k)$ for some $x_k\in [k,k+1)$, we obtain $$|a_k|\leqslant \frac 2{k^{3/2}},$$
hence $\sum_{k\geqslant 1}|a_k|$ is convergent.
