Prove that $\lim\limits_{n\to\infty}\left(\sum\limits_{r=0}^{n-1}\sqrt{1-\frac{r^2}{n^2}}-\frac{\pi}{4}n\right)=\frac{1}{2}$

Question

I came across the above question in a mathematical problem. It is not difficult to see that $$ \lim\limits_{n\to\infty}\left(\frac{1}{n}\sum\limits_{r=0}^{n-1}\sqrt{1-\frac{r^2}{n^2}}\right)=\int\limits_0^1\sqrt{1-\frac{x^2}{n^2}}\mathrm{d}x=\frac{\pi}{4}. $$ Nevertheless, the above question asks for estimating the residual of the Riemann sum. By letting $n$ become very large and computing brutely using a program, I suspect that the above limit should be $\frac{1}{2}$, but I have no idea how to prove this result mathematically. Can anybody help me?

Answer 1

By the Euler–MacLaurin summation formula for $$f(u):=f_n(u):=\sqrt{1-u^2/n^2}$$ and $m=2$ (see e.g. formulas (2.1), (2.2), and (2.4) here or here), $$S:=S_n:=\sum_{r=0}^{n-1}f(r)=A_2+R_2,$$ where \begin{align}&A_2:=A_{n;2}:= \\ &\int_0^{n-1} f(u) \, du+\frac{1}{2} (f(n-1)+f(0))+ \frac{1}{12} \left(f'(n-1)-f'(0)\right) \\ &= \frac n2\, \Big(\frac{2}{\sqrt{2 n-1}}+\pi \Big) +\frac{1}{12n \sqrt{2 n-1}}-\frac{7}{12\sqrt{2n-1}} \\ &+n \tan ^{-1}\Big(\frac{n-1}{\sqrt{2 n-1}-n}\Big)+\frac12 \end{align} and $|R_2|=O(1/\sqrt n)=o(1)$ (as $n\to\infty$). Note also that $$A_2-\frac\pi 4\,n\to\frac12.$$ So, the desired result follows.