How to prove $\lim_{x\rightarrow0^+}\sum_{k=1}^{+\infty}\frac{\sin{(x\sqrt{k})}}{k}=\pi$?

Using Ron Gordon's hint the sum tends to the integral as $x \to 0+$. An estmate of the error follows from Euler-Maclaurin:

$$\sum_{k=1}^n f(k) - \int_1^n f(y) \, dy = \frac{1}{2}(f(n) + f(1)) + \int_1^n (\{y \}- 1/2)f'(y) \, dy$$

With $f(y) = \sin (x \sqrt{y})/y$ and taking $n \to \infty$ we have

$$\sum_{k=1}^\infty \frac{\sin(x\sqrt{k})}{k} - 2\int_x^\infty \frac{\sin u}u \, du \\ = \frac{1}{2} \sin x + \int_1^\infty (\{y \}- 1/2)\left(x \frac{\cos(x\sqrt{y})}{2 y^{3/2}} - \frac{\sin(x \sqrt{y})}{y^2} \right) \, dy$$

The integral on the RHS is absolutely and uniformly convergent and the limit is $0$ as $x \to 0$. The integral on the LHS converges to $\pi$.

We may compute the givin limit through a convolution with an approximate identity:

$$ \lim_{m\to +\infty}\sum_{k\geq 1}\int_{0}^{+\infty}\frac{\sin(x\sqrt{k})}{\sqrt{k}}\sqrt{m}e^{-\sqrt{m}\,x}\,dx=\lim_{m\to +\infty}\sum_{k\geq 1}\frac{\sqrt{m}}{\sqrt{k}(k+m)}\\=\lim_{m\to +\infty}\frac{1}{m}\sum_{k\geq 1}\frac{1}{\sqrt{\frac{k}{m}}\left(\frac{k}{m}+1\right)}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x}(x+1)}=2\int_{0}^{+\infty}\frac{du}{1+u^2}=\color{red}{\pi}.$$