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Proof Of Leibniz FormulaIn 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 side with upper limit 1 and lower limit 0 -
\int_{0}^{1} ( 1 - x^2 + x^4 - x^6 + x^8 - \cdots ) dx = \int_{0}^{1} \frac{1} {1+x^2} dx The LHS becomes the required sum, -
\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} Q.E.D
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