How to prove $\lim_{x\rightarrow0^+}\sum_{k=1}^{+\infty}\frac{\sin{(x\sqrt{k})}}{k}=\pi$? $$\lim_{x\rightarrow0^+}\sum_{k=1}^{+\infty}\frac{\sin{(x\sqrt{k})}}{k}=\pi$$
I have tried to expand the $\sin{(x\sqrt{k})}$ but still cannot get $\pi$.
 A: We may compute the givin limit through a convolution with an approximate identity:
$$ \lim_{m\to +\infty}\sum_{k\geq 1}\int_{0}^{+\infty}\frac{\sin(x\sqrt{k})}{\sqrt{k}}\sqrt{m}e^{-\sqrt{m}\,x}\,dx=\lim_{m\to +\infty}\sum_{k\geq 1}\frac{\sqrt{m}}{\sqrt{k}(k+m)}\\=\lim_{m\to +\infty}\frac{1}{m}\sum_{k\geq 1}\frac{1}{\sqrt{\frac{k}{m}}\left(\frac{k}{m}+1\right)}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x}(x+1)}=2\int_{0}^{+\infty}\frac{du}{1+u^2}=\color{red}{\pi}.$$
A: 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 $\pi$.
A: (I am not sure about correctness of my approach but i am writing it here.) 
Consider the sum $S_n$ as,
$$
S_n=\sum_{k=1}^{n}\frac{\sin{(x\sqrt{k})}}{k}
$$
Now as the function $f(k) (=\frac{\sin{(x\sqrt{k})}}{k})$ is decreases with $k$ (if $x$ is positive) (see note below). So,
we can use lower and upper bounds as,
$$
\int_1^{\infty} f(k) \, dk \leq S_{\infty} \leq f(1)+\int_1^{\infty} f(k) \, dk
$$
But $f(1)=sin(x)$ so,
$$
\int_1^{\infty} \frac{\sin{(x\sqrt{k})}}{k} \, dk \leq S_{\infty} \leq sin(x)+\int_1^{\infty} \frac{\sin{(x\sqrt{k})}}{k} \, dk
$$
Taking $\lim_{x\rightarrow0^+}$ on this we get,
$$
\lim_{x\rightarrow0^+} \int_1^{\infty} \frac{\sin{(x\sqrt{k})}}{k} \, dk \leq \lim_{x\rightarrow0^+}S_{\infty} \leq \lim_{x\rightarrow0^+}\int_1^{\infty} \frac{\sin{(x\sqrt{k})}}{k} \, dk
$$
Hence,
$$
\lim_{x\rightarrow0^+}S_{\infty} = \lim_{x\rightarrow0^+}\int_1^{\infty} \frac{\sin{(x\sqrt{k})}}{k} \, dk
$$
Put $t=x\sqrt{k}$ and solve to get,
$$
\lim_{x\rightarrow0^+}S_{\infty} = \lim_{x\rightarrow0^+} 2\int_x^{\infty} \frac{\sin{(t)}}{t} \, dt
$$
And so
$$
\lim_{x\rightarrow0^+}S_{\infty} = 2\lim_{x\rightarrow0^+} \int_x^{\infty} \frac{\sin{(t)}}{t} \, dt = 2 \int_0^{\infty} \frac{\sin{(t)}}{t} \, dt = 2 \frac{\pi}{2} = \pi
$$
Note:
I can say the function $f(k)=\frac{\sin{(x\sqrt{k})}}{k}$ is a decreasing function as, Let say if $0\leq x\sqrt{k} \leq \frac{\pi}{2}$ to avoid periodicity in $sine$ then $0\leq k \leq \frac{\pi^2}{4x^2}$. Take $\lim_{x\rightarrow0^+}$ and hence $0\leq k < \infty$ which is true. And so we can say that $f(k)$ is decresing function with ${x\rightarrow0^+}$. 
