Need to simplify formula $ \pi \sim \frac{8}{n^2} \cdot \sum\limits_{i=0}^{n} \sqrt{i \cdot (n - i)} $

Question

This formula can be simplified?

$$ \pi \sim \frac{8}{n^2} \cdot \sum\limits_{i=0}^{n} \sqrt{i \cdot (n - i)} $$

I am trying to find an alternative formula for circle area.
I stacked to analyze this formula for a while without any progress.

This is not my home work.
I am asking here, because I know all the professional mathematician are here and maybe exist such theorem that I do not know yet and can help to move further with my research.

Thank You.

I tested convergence numerically with:

import math

p = 6
n = 10 ** p

sum_h = 0
for i in range(n+1):
    h = (i * (n - i)) ** (1/2)
    sum_h = sum_h + h

pi0 = 8 * sum_h / n / n

print(" ", math.pi)
print("~", pi0)

  3.141592653589793
~ 3.141592650263593

Here is what I know.

Preview integral example.

We are so close!

Answer 1

@Gary gave in comments the most perfect asymptotic formula you could expect.

Working the asymptotics of a lower bound, consider $$F(n)=\sum\limits_{i=0}^{n} \sqrt{i\,(n - i)}\qquad\text{and}\qquad G(n)=\int_0^{n-1} \sqrt{i\,(n - i)}\,di$$ $$G(n)=\frac{1}{4} \left(n^2 \tan ^{-1}\left(\sqrt{n-1}\right)+(n-2) \sqrt{n-1}\right)$$ For all $n$ the sum is larger than the integral. Recall that $$\int_0^{n} \sqrt{i\,(n - i)}\,di=\frac{\pi }{8}n^2$$

Now, using series expansion, the lower bound is $$G(n)=\frac{\pi }{8}n^2-\frac{2 \sqrt{n}}{3}\left( 1-\frac{3}{10 n}-\frac{3}{56 n^2}-\frac{1}{48 n^3}+O\left(\frac{1}{n^4}\right)\right)$$ which makes $$\frac 8{n^2}F(n)> \pi-\frac{16}{3 n^{3/2}}\left( 1-\frac{3}{10 n}-\frac{3}{56 n^2}-\frac{1}{48 n^3}+O\left(\frac{1}{n^4}\right)\right)$$

For $n=100$, the lhs is $3.13827$ while the rhs is $3.13628$

Answer 2

Provided formula $$ \pi \sim \frac{8}{n^2} \cdot \sum\limits_{i=0}^{n} \sqrt{i \cdot (n-i)} $$ is similar with $$ \pi \sim \frac{4}{n^2} \cdot \sum\limits_{i=0}^{n} \sqrt{n^2 - i^2} = \frac{4}{n^2} \cdot \sum\limits_{i=0}^{n} \sqrt{(n+i) \cdot (n-i)} $$

Expressing equivalently with integral

$$ \int_{0}^{n} \sqrt{n-x} \sqrt{n+x} dx = \frac{\pi n^2}{4} $$

Answer 3

Using binomial expansion:

$$\int\limits_{0}^{1} \sqrt{1-x^2} dx = \sum\limits_{k=0}^{\infty} (-1)^k \frac{(2k-1)!!}{(2k)!!} \int\limits_{0}^{1} x^{2k}dx$$ $$\int\limits_{0}^{1} \sqrt{1-x^2} dx = \sum\limits_{k=0}^{\infty} (-1)^k \frac{(2k-1)!!}{(2k)!!} \frac{1}{2k+1} \to \frac{\pi}{4}$$