How to compute $\lim_{n\to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-k^2}}$

Question

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

Answer 1

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.

$$ 0 \cdot \int_{0}^{1} \frac{1}{ \sqrt{(1+x^2)(1-x^2)} \sqrt { (1-x^2) }+\sqrt{(1+x^2)}}\rm{d}x=0$$

Answer 2

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.

Answer 3

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 $\frac\pi2 \leq I \leq \frac\pi2$.

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $\approx -0.00135758$.