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 $$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\le \frac{1}{1+(k/n)^2}$$
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 $\frac{n}{n^2+k^2}\le\frac1n$, $$ \begin{align} \sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\!\right) &=\sum_{k=1}^n\frac{n}{n^2+k^2}-\sum_{k=1}^n\left[\frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\!\right)\right]\\ &=\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\frac1n-\sum_{k=1}^nO\!\left(\frac1{n^3}\right)\\ &\to\int_0^1\frac1{1+x^2}\,\mathrm{d}x-0\\[6pt] &=\frac\pi4 \end{align} $$
Use Euler-Maclaurin to approximate the sum as $$ \int_0^n \sin\left(\frac{n}{n^2+x^2}\right)dx\ . $$ Then change variables $x=ny$ and use $\sin(\alpha/n)\sim\alpha/n$ for $n\to\infty$, to obtain the result $\int_0^1 dy\frac{1}{1+y^2}=\pi/4$.
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 $$ \lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)= \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{\pi}{4}. $$
