How to calculate $\lim_{n \to \infty} \sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}$? How to calculate $\displaystyle\lim_{n \to \infty} \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}$?
My try:
\begin{align}
\lim_{n \to \infty} \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}
&=\displaystyle \lim_{n \to \infty} \frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {1+\frac{1}{n}-(\tfrac{k}{n})^2}}
\\&=\displaystyle\lim_{n \to \infty}\int_{0}^{1}\frac{dx}{\sqrt{1+\frac{1}{n}-x^2}}
\\&=\displaystyle\lim_{n \to \infty}\arctan \sqrt{n}
\\&=\frac{\pi}{2}
\end{align}
But,
I'm not sure whether this's right because I'm not sure whether the second equality is right.
Any helps and new ideas will be highly appreciated!
 A: $$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt {1+\epsilon-(\tfrac{k}{n})^2}}  \preceq \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt {1+\frac{1}{n}-(\tfrac{k}{n})^2}} \leq \frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {1-(\tfrac{k}{n})^2}} $$
(the symbole  $\preceq$ means: It is lower than form a $n\in \mathbb{N}$ to later)
But $$\lim_{n\to \infty} \frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {1-(\tfrac{k}{n})^2}}=\int_0^1\arcsin(x)dx=\frac{\pi}{2}  $$ and $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt {1+\epsilon-(\tfrac{k}{n})^2}} =\arcsin \left(\frac{1}{\sqrt {1+\epsilon}}\right)$$ and 
$$\lim_{\epsilon \to 0^+} \arcsin\left(\frac{1}{\sqrt {1+\epsilon}}\right)=\frac{\pi}{2}.$$
A: To use integral method rigorously, I came up with a new solution.
Notice that(due to the monotonicity)
$$ \displaystyle\int_{0}^{n}\frac{dx}{\sqrt {n^2+n-x^2}} \le \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}\le\int_{1}^{n}\frac{dx}{\sqrt {n^2+n-x^2}}+\frac{1}{\sqrt{n}}$$
Then we have
$$\displaystyle\lim_{n\to\infty}\displaystyle\int_{0}^{n}\frac{dx}{\sqrt {n^2+n-x^2}} \le \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}\le\displaystyle\lim_{n\to\infty}\int_{1}^{n}\frac{dx}{\sqrt {n^2+n-x^2}}$$
Considering 
$$\displaystyle\int\frac{dx}{\sqrt {n^2+n-x^2}}=\arctan\frac{x}{\sqrt {n^2+n-x^2}}$$
Then we can arrive at 
$$\displaystyle\lim_{n \to \infty} \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}=\frac{\pi}{2}$$
