I found that the series
$$s(n) = \frac{2}{n} \cdot \sum_{i=1}^n \sqrt{\frac{n}{i-\frac{1}{2}}-1}$$
converges to $\pi$ as $n \to \infty$.
To verify this I have computed some values:
| $n$ | $s(n)$ |
|---|---|
| $10^1$ | 2.76098 |
| $10^3$ | 3.10333 |
| $10^5$ | 3.13776 |
| $10^6$ | 3.14038 |
| $10^7$ | 3.14121 |
Which seems to support the claim, however, this is no proof of the convergence.
I would not know how to begin on a proof of this limit and did not find any similar formula in known approximation formulas.
Does anyone have an idea on how such a proof can be constructed?
Best Answer
$\newcommand{\d}{\,\mathrm{d}}$Note that: $$\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{1}{k/n-1/2n}-1}\to\int_0^1\sqrt{x^{-1}-1}\d x=\frac{\pi}{2}$$
Substituting $x=t^{-1},t=u+1,u=w^2$ in that order equates the integral with: $$2\int_0^\infty\frac{w^2}{(w^2+1)^2}\d w$$And this is manageable.
Or, let $x=\sin^2t$. You then deal with: $$\int_0^{\pi/2}2\sin(t)\cos(t)\cot(t)\d t=2\int_0^{\pi/2}\cos^2t\d t$$Which is even easier.
To justify treating the partial sums as Riemann sums, it is sufficient to demonstrate: $$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left[\left(\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k-\frac{1}{2}}-1}\right)-\left(\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k}-1}\right)\right]=0$$
I present a proof of this by elementary bounds. No need for fancy convergence theorems here!
Using the difference of two squares we can bound, for every $k\ge1$ and every $n>k$: $$\begin{align}0&\le\sqrt{\frac{n}{k-\frac{1}{2}}-1}-\sqrt{\frac{n}{k}-1}\\&=\frac{\frac{\frac{1}{2n}}{\frac{k}{n}\left(\frac{k}{n}-\frac{1}{2n}\right)}}{\sqrt{\frac{n}{k-\frac{1}{2}}-1}+\sqrt{\frac{n}{k}-1}}\\&\le\frac{1}{2}\sqrt{\frac{k}{n-k}}\cdot\frac{n}{k(2k-1)}\\&=\frac{1}{2}\frac{\sqrt{n}}{(2k-1)\sqrt{k(1-k/n)}}\\&\le\frac{\sqrt{n}}{2k\sqrt{k(1-k/n)}}\end{align}$$Summing in $k$ and dividing by $n$ finds: $$0\le S_n\le\frac{1}{n\sqrt{2n-1}}+\frac{1}{2\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}$$
Estimating the sum explicitly is a little fiddly. The map $x\mapsto\frac{1}{x\sqrt{x(1-x/n)}}$ is convex. It is initially decreasing but then increases after $x=3n/4$. If $m$ is the floor of $3n/4$ and $n$ is considered large, we can put: $$\begin{align}\sum_{k=1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}&=\sum_{k=1}^m\frac{1}{k\sqrt{k(1-k/n)}}+\sum_{k=m+1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}\\&\le\frac{1}{\sqrt{1-1/n}}+\int_1^m\frac{1}{x\sqrt{x(1-x/n)}}\d x\\&+\int_{m+1}^{n-1}\frac{1}{x\sqrt{x(1-x/n)}}\d x+\frac{1}{(n-1)\sqrt{1-1/n}}\\&<(1-1/n)^{-3/2}+\int_1^{n-1}\frac{1}{x\sqrt{x(1-x/n)}}\d x\\&=(1-1/n)^{-3/2}+\frac{2(n-2)}{\sqrt{n(n-1)}}\end{align}$$
Overall I get: $$0<S_n<\frac{1}{n\sqrt{2n-1}}+\frac{n}{2(n-1)\sqrt{n-1}}+\frac{n-2}{n\sqrt{n-1}}$$When $n$ is large enough that $m<n-2$, e.g. this is for sure when $n>12$.
By the squeeze theorem, it is now clear that $S_n$ vanishes - as desired.