How to compute $\lim_{n\to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-k^2}}$ Find  this follow limit
$$I=\lim_{n\to \infty}\sum_{k=1}^{n}\dfrac{1}{\sqrt{n^2+n-k^2}}$$
since
$$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}$$
I guess  we have
$$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1-(k/n)^2}}=\int_{0}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{\pi}{2}$$
But I can't prove  follow is true
$$\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1-(k/n)^2}}$$
I have only prove 
$$\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}<\dfrac{1}{\sqrt{1-\left(\dfrac{k}{n}\right)^2}}$$
Thank you

You have already establised that $I \leq \frac\pi2$.
If you instead divide numerator and denominator by $n+1$ rather than $n$ you get:
$$ I = \lim_{n\to\infty} \frac1{n+1} \sum_{k=1}^n \frac{1}{\sqrt{1 - \frac1{n+1} - \left(\frac{k}{n+1}\right)^2}} $$
Letting $m = n+1$, and adding and subtracting the $0$th summand this is:
$$ \begin{aligned} I 
 & = \lim_{m\to \infty} \left( \frac1m \sum_{k=0}^{m-1} \frac{1}{\sqrt{1-\frac1m - \left(\frac k m\right)^2}} - \frac1m \frac1{\sqrt{1 - \frac1m - \left(\frac 0m\right)^2}} \right) \\ 
 & = \lim_{m\to \infty} \left( \frac1m \sum_{k=1}^m \frac{1}{\sqrt{1-\frac1m - \left(\frac k m\right)^2}} - \frac1m \frac1{\sqrt{1 - \frac1m }} \right) \\
 & = \lim_{m\to \infty} \left( \frac1m \sum_{k=1}^m \frac{1}{\sqrt{1-\frac1m - \left(\frac k m\right)^2}}\right) - \lim_{m\to \infty} \left(\frac1m \frac1{\sqrt{1 - \frac1m }} \right) 
\end{aligned} $$
The second limit is $0$.  The first limit is bounded below by the integral $\frac\pi2$ that you found, just as in your limit you found it was bounded above.  Thus $\frac\pi2 \leq I \leq \frac\pi2$.
