User:Constan69

I am me.

In mathematics, Leibniz' formula for π, due to Gottfried Leibniz, states that

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

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 $|x|<1$)

$\arctan x=\sum _{n\geq 0}(-1)^{n}{x^{2n+1} \over {2n+1}}$

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.