TITLE: Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
QUESTION [8 upvotes]: Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$

$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\sin\left(\frac{n}{n^2+r^2}\right)$$
Now how can I convert into Riemann Sum, Help me
Thanks

REPLY [3 votes]: For $x\gt0$, repeatedly integrating from $0$ to $x$ gives
$$
\cos(x)\le1\implies\sin(x)\le x\implies1-\cos(x)\le\frac{x^2}2\implies x-\sin(x)\le\frac{x^3}6
$$
Noting that both sides are odd, we get
$$
\left|x-\sin(x)\right|\le\frac{\left|x^3\right|}6
$$
Since $\frac{n}{n^2+k^2}\le\frac1n$,
$$
\begin{align}
\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\!\right)
&=\sum_{k=1}^n\frac{n}{n^2+k^2}-\sum_{k=1}^n\left[\frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\!\right)\right]\\
&=\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\frac1n-\sum_{k=1}^nO\!\left(\frac1{n^3}\right)\\
&\to\int_0^1\frac1{1+x^2}\,\mathrm{d}x-0\\[6pt]
&=\frac\pi4
\end{align}
$$