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?
 A: If you like generalized harmonic numbers, use the Taylor series of $\sin(x)$.
Now, write
$$x=\frac{n}{k^2+n^2}=\frac{n}{(k+in) (k-in)}=\frac i 2\Big[\frac 1{k+i n}-\frac 1{k-i n} \Big]$$ DO the same for $x^3$ and use partial fracion decomposition. Compute the summation for $k=n+1$ to $k=2n$ and use the asymptotics of the generated generalized harmonic numbers. You should arrive at
$$\sum _{k=n+1}^{2 n} \sin \left(\frac{n}{k^2+n^2}\right)=\frac{1}{4} \left(\pi -4 \cot ^{-1}(2)\right)-\frac{3}{20 n}+\frac{8-3 \pi +12 \cot
   ^{-1}(2)}{192 n^2}+\frac{39}{4000 n^3}+O\left(\frac{1}{n^4}\right)$$ and notice that the constant term is also $\tan ^{-1}\left(\frac{1}{3}\right)$ as already given by @metamorphy.
Trying for $n=10$, the decimal representation of the sum is $\color{red}{0.306975}42$ while the truncated series gives
$$\frac{-174883+2998125 \pi -11992500 \cot ^{-1}(2)}{12000000}=\color{red}{0.30697588}$$
A: $$S_n:=\sum_{k=n+1}^{2n}\sin\frac{n}{n^2+k^2},\quad T_n:=\sum_{k=n+1}^{2n}\frac{n}{n^2+k^2},\quad\lim_{n\to\infty}T_n=\int_1^2\frac{dx}{1+x^2}$$ (using Riemann sums for the last one) and, since $0\leqslant x-\sin x\leqslant x^3/6$ for $x\geqslant 0$, we have $$0\leqslant T_n-S_n\leqslant\frac16\sum_{k=n+1}^{2n}\left(\frac{n}{n^2+k^2}\right)^3\leqslant\frac16\cdot n\cdot\frac1{n^3}\underset{n\to\infty}{\longrightarrow}0.$$ Hence $\lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}T_n\color{Violet}{=\tan^{-1}(1/3)}$.
A: Can't we use the approximation $\sin(\frac{n}{n^2+k^2})\approx \frac{n}{n^2+k^2}$ when $n$ is big?
Then it is the Rieamann sum WE ALL KNOW VERY WELL:
$$\sum_{k=n+1}^{2n}\frac{n}{n^2+k^2}=\sum_{k=n+1}^{2n}\frac{1}{n}\frac{1}{1+(k/n)^2}=\int_1^2\frac{1}{1+x^2}dx=\tan^{-1}(2)-\frac{\pi}{4}=\tan^{-1}(2)-\tan^{-1}(1)=\boxed{\color{blue}{\tan^{-1}\left(\frac13\right)}}\approx 0.32175$$
My approximation is justified here: https://www.wolframalpha.com/input?i=sum+10000%2F%2810000%5E2%2Bk%5E2%29+%2C+k%3D10001+to+20000
And here: https://www.wolframalpha.com/input?i=sum+sin%2810000%2F%2810000%5E2%2Bk%5E2%29%29+%2C+k%3D10001+to+20000
A: 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
$$\frac{4-70 \pi +40 \cot ^{-1}(2)}{125 \pi ^2\,n}\times\frac {20}3\sim\frac {16}{15}$$
