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.
Best Answer
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.