Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$ 
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

We can use Riemann sums to evaluate the limit. We have
$$\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=\sum_{k=1}^n\,n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\frac1n\to \int_0^1\frac{1}{1+x^2}\,dx=\pi/4$$
since we have $$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\le \frac{1}{1+(k/n)^2}$$
