User:Constan69

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In mathematics, Leibniz' formula for π, due to Gottfried Leibniz, 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}.

[edit] Proof

Consider the infinite geometric series

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

It is the limit of the truncated geometric series

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

Splitting the integrand as

\frac{1} {1+x^2}=\frac{1-x^{4n}}{1+x^2}+\frac{x^{4n}}{1+x^2}=G_n (x)+ \frac{x^{4n}}{1+x^2}

and integrating both sides from 0 to 1, we have

\int_{0}^{1} \frac{1} {1+x^2}\, dx= \int_{0}^{1}G_n(x)\, dx+\int_{0}^{1}\frac{x^{4n}}{1+x^2}\, dx \ .

Integrating the first integral (over the truncated geometric series G_n (x)\,) termwise one obtains in the limit the required sum. The contribution from the second integral vanishes in the limit n \rightarrow \infty as

\int_{0}^{1}\frac{x^{4n}}{1+x^2} \, dx< \int_{0}^{1} x^{4n}\, dx=\frac{1}{4n+1} \ .

The full integral

\int_{0}^{1} \frac{1} {1+x^2}\, dx

on the left-hand side evaluates to arctan(1) − arctan(0) = π/4, which then yields

\frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots.

Q.E.D.

Remark: An alternative proof of the Leibniz formula can be given with the aid of Abel's theorem applied to the power series (convergent for | x | < 1)

\arctan x =\sum_{n \ge 0} (-1)^n {x^{2n+1}\over {2n+1}}

which is obtained integrating the geometric series ( absolutely convergent for | x | < 1)

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

termwise.

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