Evaluating $\lim _{n \rightarrow \infty} \sum_{k=n+1}^{2 n} \sin \frac{n}{n^{2}+k^{2}}$ Evaluate $$S=\lim _{n \rightarrow \infty} \sum_{k=n+1}^{2 n} \sin \frac{n}{n^{2}+k^{2}}$$
My try:
Since $\sin x$ is monotone increasing in $(0,1)$, we have:
$$n \sin \left (\frac{n}{n^2+(2n)^2}\right)\leq S \leq n\sin \left(\frac{n}{n^2+(n+1)^2}\right)$$
taking $\lim$ throughout we get:
$$0.2 \leq S \leq 0.5$$
I am stuck now?

Another answer (just for the fun of it)
Using  my favored $\large 1,400$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ make $x=\frac{n}{k^2+n^2}$ and compute
$$S_n=\frac{16 n}{5 \pi }\sum _{k=n+1}^{2 n} \frac{k^2+\alpha }{\left(k^2+\beta \right) \left(k^2+\gamma \right)}$$  where
$$\alpha=n \left(n-\frac{1}{\pi }\right)$$
$$\beta=n^2-\frac{\left(\frac{2}{5}-\frac{4 i}{5}\right) n}{\pi }\qquad \qquad \gamma=n^2-\frac{\left(\frac{2}{5}+\frac{4 i}{5}\right) n}{\pi }$$ This will give a bunch of polylogarithms.
Expanding as a series for large values of $n$, the constant term is
$$\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }$$ and comparing to the exact result
$$\frac{\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }} {\frac{1}{4} \left(\pi -4 \cot ^{-1}(2)\right)}=\frac{16}{5 \pi }\sim \frac{56}{55}\qquad \text{using} \qquad \pi\sim \frac{22}{7}$$
The asymptotics is
$$\sum _{k=n+1}^{2 n} \sin \left(\frac{n}{k^2+n^2}\right)=\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }-\frac{4-70 \pi +40 \cot ^{-1}(2)}{125 \pi ^2\,n}+$$ $$\frac{2 \left(5 \pi  (85 \pi -207)+72 \left(16+75 \cot ^{-1}(2)\right)\right)}{9375
   \pi ^3\,n^3}+O\left(\frac{1}{n^3}\right)$$ and notice again that
$$\frac{4-70 \pi +40 \cot ^{-1}(2)}{125 \pi ^2\,n}\times\frac {20}3\sim\frac {16}{15}$$
