𝜋 Approximation (#91)

Jonathan Frech, 19 December 2015

Using an infinite series shown by Euler, 𝜋 can be approximated.
The series goes as follows: $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots=\frac{\pi^2}{6}$ $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots=\frac{\pi^2}{6}$
By rearranging the equation, you get the following: $\pi=\sqrt{6\cdot\big(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots\big)}$ $\pi=\sqrt{6\cdot\big(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots\big)}$

Source code: pi-approximation.py