<s>[INST] 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
 [/INST] 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. 
$$ 0 \cdot \int_{0}^{1} \frac{1}{  \sqrt{(1+x^2)(1-x^2)}  \sqrt { (1-x^2)   }+\sqrt{(1+x^2)}}\rm{d}x=0$$
</s>