compute $\lim_{n\rightarrow\infty}\sum_{k=1}^n \sin(\pi \sqrt{k/n}) (1/\sqrt{kn})$

I'm currently studying for an analysis exam and encountered this problem on an old exam: Calculate the limit: $$\displaystyle{\lim_{n \rightarrow \infty} \sum_{k=1}^n \sin \left ( \pi \sqrt{\frac{k}{n}} \right ) \left ( \frac{1}{ \sqrt{kn}} \right ) }$$ The exam solutions are not included, and I'm at a loss as to how to approach this one.

Thanks in advance for the help!

$$\lim_{n \to \infty} \sum_{k =1 }^{n} \sin \Bigg(\pi \sqrt{\frac{k}{n}}\Bigg)\Bigg(\frac{1}{\sqrt{kn}}\Bigg) \underbrace{=}_{(*)} 2\int_{0}^{1} \sin \pi x \,dx = \color{red}{\frac{4}{\pi}}$$

Spoiler for $(*)$:

\begin{align}\lim_{n \to \infty} \sum_{k =1 }^{n} \sin \Bigg(\pi \sqrt{\frac{k}{n}}\Bigg)\Bigg(\frac{1}{\sqrt{kn}}\Bigg) &= \lim_{n \to \infty} \sum_{k =1 }^{n} \sin \Bigg(\pi \sqrt{\frac{k}{n}}\Bigg)\Bigg(\frac{n}{\sqrt{kn}}\Bigg)\Bigg(\frac{1}{n}\Bigg) \\&= \lim_{n \to \infty} \sum_{k =1 }^{n} \sin \Bigg(\pi \sqrt{\frac{k}{n}}\Bigg)\Bigg(\sqrt{\frac{n}{k}}\Bigg)\Bigg(\frac{1}{n}\Bigg) \\&=\int_{0}^{1} \frac{\sin \pi \sqrt{t}}{\sqrt{t}} dt \end{align} Take $\Delta t= \frac{1}{n}$, $a = 0$ and notice that $$\lim_{n \to \infty} \sum_{k =1 }^{n} f(a + k \Delta t)\Delta t = \int_a^b f(t)dt$$ then $\frac{1}{n} = \frac{b - a}{n} \implies b = 1$. Take $x =\sqrt{t}$ then $dx = \frac{1}{2\sqrt{t}}dt$ and you get $(*)$.

• I believe you forgot a factor. – Lucian Apr 20 '15 at 23:15
• You're right, thanks. – Aaron Maroja Apr 20 '15 at 23:23
• The $\frac1n$ in the spoiler disappears. – robjohn Apr 20 '15 at 23:54
• @robjohn thanks, man. – Aaron Maroja Apr 20 '15 at 23:56
• Thanks, Aaron! Chose this as the answer because the detail contained in the spoiler was particularly helpful. – user218389 Apr 21 '15 at 3:25

Multiplying by $\dfrac nn$ we have $~\displaystyle\lim_{n\to\infty}~\frac1n~\sum_{k=1}^n\frac{\sin\bigg(\pi~\sqrt{\dfrac kn}~\bigg)}{\sqrt{\dfrac kn}},~$ which is the Riemann sum for

$\displaystyle\int_0^1\frac{\sin\pi\sqrt x}{\sqrt x}~dx,~$ which, after letting $t=\sqrt x,~$ becomes $~\color{red}2\displaystyle\int_0^1\sin\pi t~dt,~$ whose evaluation

is trivial.

• I believe you miss a factor $\frac{1}{n}$ in your Riemann sum. – Guest Apr 20 '15 at 23:25