TITLE: How to prove $\lim_{x\rightarrow0^+}\sum_{k=1}^{+\infty}\frac{\sin{(x\sqrt{k})}}{k}=\pi$?
QUESTION [11 upvotes]: $$\lim_{x\rightarrow0^+}\sum_{k=1}^{+\infty}\frac{\sin{(x\sqrt{k})}}{k}=\pi$$
I have tried to expand the $\sin{(x\sqrt{k})}$ but still cannot get $\pi$.

REPLY [5 votes]: We may compute the givin limit through a convolution with an approximate identity:
$$ \lim_{m\to +\infty}\sum_{k\geq 1}\int_{0}^{+\infty}\frac{\sin(x\sqrt{k})}{\sqrt{k}}\sqrt{m}e^{-\sqrt{m}\,x}\,dx=\lim_{m\to +\infty}\sum_{k\geq 1}\frac{\sqrt{m}}{\sqrt{k}(k+m)}\\=\lim_{m\to +\infty}\frac{1}{m}\sum_{k\geq 1}\frac{1}{\sqrt{\frac{k}{m}}\left(\frac{k}{m}+1\right)}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x}(x+1)}=2\int_{0}^{+\infty}\frac{du}{1+u^2}=\color{red}{\pi}.$$