Prove-that-pi-4-k-0-1-k-2k-1-

Question Number 7621 by 314159 last updated on 06/Sep/16

P r o v e t h a t

π = 4 \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{2 k + 1}

Answered by prakash jain last updated on 26/Sep/16

Outline for the proof :

x < 1

\frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - x^{6} + x^{8} - + . .

Integrate from 0 to 1 .

\int_{0}^{1} \frac{1}{1 + x^{2}} = \int_{0}^{1} 1 - x^{2} + x^{4} - x^{6}

{[\tan^{- 1} x]}_{0}^{1} = {[x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + - . .]}_{0}^{1}

\frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + - \dots

\frac{π}{4} = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{2 k + 1}