User:Constan69

I am me.

In mathematics, Leibniz' formula for &pi;, 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}.

$$

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) &minus; arctan(0) = &pi;/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.