sociology - Proof of Leibniz formula

In mathematics, Leibniz' formula for π states that

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \frac{\pi}{4}.$

Proof

Consider the infinite geometric series

$1 - x^2 + x^4 - x^6 + x^8 - \cdots = \frac{1}{1+x^2}, \qquad |x| < 1.$

Integrating both sides from 0 to 1, we have

$\int_{0}^{1} ( 1 - x^2 + x^4 - x^6 + x^8 - \cdots )\, dx = \int_{0}^{1} \frac{1} {1+x^2}\, dx.$

The left-hand side then becomes the required sum, while the right-hand side evalulates to π/4:

$\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \tan^{-1} (1) - \tan^{-1} (0) = \frac{\pi}{4}.$