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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!

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  • $\begingroup$ As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct. $\endgroup$ – Will M. Dec 21 '18 at 5:39
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$$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt {1+\epsilon-(\tfrac{k}{n})^2}} \preceq \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt {1+\frac{1}{n}-(\tfrac{k}{n})^2}} \leq \frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {1-(\tfrac{k}{n})^2}} $$ (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 $$\lim_{\epsilon \to 0^+} \arcsin\left(\frac{1}{\sqrt {1+\epsilon}}\right)=\frac{\pi}{2}.$$

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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 $$\displaystyle\lim_{n \to \infty} \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt {n^2+n-k^2}}=\frac{\pi}{2}$$

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