<s>[INST] 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
 [/INST] As a consequence of the fact that $\,\,\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\big(\frac{k}{n}\big)$, we obtain
$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\to\int_0^1\frac{dx}{1+x^2}=\tan^{-1}1=\frac{\pi}{4}
$$
Next, Taylor expansion of $\sin x$, provides that $0\le x-\sin x\le \frac{x^3}{6} $, for $0\le  x\le 1$.
Thus
$$
0\le \frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\right)\le \frac{1}{6n^3},
$$
and thus
$$
0\le \sum_{k=1}^n\left(\frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\right)\right)\le \frac{1}{6n^2}.
$$
Thus
$$
\lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=
\lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{\pi}{4}.
$$
</s>