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
 A: The difference term-wise is 
\begin{align}
\frac{1}{ \sqrt {1+({\frac{k}{n}})^2}} - \frac{1}{ \sqrt {1 + \frac{1}{n} - {(\frac{k}{n}})^2}}&= \frac{\frac{1}{n}}{    \sqrt {1+({\frac{k}{n}})^2}\cdot  \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2}  ( \sqrt {1+({\frac{k}{n}})^2}+  \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2})  }  \\
&\leq \frac{1}{n}\frac{1}{   \sqrt {1+({\frac{k}{n}})^2}\cdot  \sqrt {1- ({\frac{k}{n}})^2}  ( \sqrt {1+({\frac{k}{n}})^2}+  \sqrt {1  - ({\frac{k}{n}})^2})     }
\end{align}
Therefore when we take the difference of the sums we get one extra $\frac{1}{n}$ and the other quantity converges to an integral thus the result is zero which proves your claim. 
$$ 0 \cdot \int_{0}^{1} \frac{1}{  \sqrt{(1+x^2)(1-x^2)}  \sqrt { (1-x^2)   }+\sqrt{(1+x^2)}}\rm{d}x=0$$
A: For any fixed $\varepsilon>0$, you  can write $\frac{1}{\sqrt{1+\varepsilon-(k/n)^2}}<\frac{1}{\sqrt{1+(1/n)-(k/n)^2}}$ for $n>n_0.$
This will converge to $$\int_0^1\frac{1}{\sqrt{1+\varepsilon-x^2}}dx.$$
Therefore, 
$$\int_0^1\frac{1}{\sqrt{1+\varepsilon-x^2}}dx \le \lim I \le \int_0^1\frac{1}{\sqrt{1-x^2}}dx,$$
 but $\varepsilon>0$ was arbitrary.
A: 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 $\frac\pi2 \leq I \leq \frac\pi2$.
A: $\newcommand{\+}{^{\dagger}}%
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Let's check what Euler-Maclaurin says:
\begin{align}
&\color{#0000ff}{\large\sum_{k = 1}^{n}{1 \over \root{n^{2} + n - k^{2}}}}={1 \over \root{n}} +
\sum_{k = 1}^{n - 1}{1 \over \root{n^{2} + n - k^{2}}}
\\[3mm]&=
{1 \over \root{n}} + \overbrace{\int_{0}^{n}{\dd k \over \root{n^{2} + n - k^{2}}}}^{\ds{\arcsin\pars{n \over\root{n^{2} + n}}}}
-
\half\pars{{1 \over \root{n^{2} + n}} + {1 \over \root{n}}}
+ {1 \over 12}\,{1 \over \root{n}}\\[3mm]&
- {1 \over 720}\pars{{15 \over \root{n}} + {9 \over n\root{n}}} + \cdots
\color{#0000ff}{\large\stackrel{n \to \infty}{\to} {\pi \over 2}}
\end{align}
It seems $\pi/2$ is the correct result but the convergence, in the numerical side, is very slow. Even with $n = 10^{8}$, the difference respect of $\pi/2$ is $\approx -0.00135758$.
