Find: $\lim\limits_{n\to \infty}\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$ Let
$$x_n=\arcsin \frac {1}{\sqrt{n^2+1}}+\arcsin \frac {1}{\sqrt{n^2+2}}+...+\arcsin \frac {1}{\sqrt{n^2+n}}$$ 
I would like to find $\lim\limits_{n\to \infty}x_n$ 

I attempted to use Riemann rectangle formula but it does not work.

Any Hint? 

for any $k\in\mathbb{N}$ we have
$$\arcsin \frac {1}{\sqrt{n^2+1}}\sim \frac{1}{n},\;n\to\infty$$
Indeed 
$$\lim_{n\to \infty } \, \frac{ \arcsin \frac{1}{\sqrt{k+n^2}}}{\frac{1}{n}}=^{(*)}\lim_{n\to \infty } \, \frac{-\frac{n}{\sqrt{(k+n^2)^3} \sqrt{1-\frac{1}{k+n^2}}}}{-\frac{1}{n^2}}=\\=\lim_{n\to \infty } \,\frac{n^3}{\sqrt{k^3+3 k^2 n^2+3 k n^4+n^6}\sqrt{1-\frac{1}{k+n^2}}}=\\=\lim_{n\to \infty } \,\frac{n^3}{n^3\sqrt{\frac{k^3}{n^6}+\frac{3k^2}{n^4}+\frac{3k}{n^2}+1}\sqrt{1-\frac{1}{k+n^2}}}=1$$
$(*)$ L'Hopital rule
Thus $$\sum_{k=1}^n\arcsin \frac {1}{\sqrt{n^2+k}} \sim n\cdot \frac{1}{n}=1$$
Therefore $$\color{red}{\lim_{n\to\infty}\sum_{k=1}^n\arcsin \frac {1}{\sqrt{n^2+k}} =1}$$
Hope this helps
