I am me.
In mathematics, Leibniz' formula for π, due to Gottfried Leibniz, states that
![{\displaystyle \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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90d4d067e4681097ec1da055bd4c7820eaebe347)
Proof
Consider the infinite geometric series
![{\displaystyle 1-x^{2}+x^{4}-x^{6}+x^{8}-\cdots ={\frac {1}{1+x^{2}}},\qquad |x|<1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fc56724ef514e520c4ba7faba61d0d560af392c)
It is the limit of the truncated geometric series
![{\displaystyle 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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d133214c873baa4dcb9127109e5175aa4ebf19b9)
Splitting the integrand as
![{\displaystyle {\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74a59f3209d52e7653f24e28e9078e48d0271c84)
and integrating both sides from 0 to 1, we have
![{\displaystyle \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\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39a37c845aeb9368d3a640b43d256ae15282d6d8)
Integrating the first integral (over the truncated geometric series
) termwise one obtains in the limit the required sum. The contribution from the second integral vanishes in the limit
as
![{\displaystyle \int _{0}^{1}{\frac {x^{4n}}{1+x^{2}}}\,dx<\int _{0}^{1}x^{4n}\,dx={\frac {1}{4n+1}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb274584118cf37b34a492ccdffb2da67729c9d2)
The full integral
![{\displaystyle \int _{0}^{1}{\frac {1}{1+x^{2}}}\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f12dc3e25111801315765e3f6009f07bb57a37e)
on the left-hand side evaluates to arctan(1) − arctan(0) = π/4, which then yields
![{\displaystyle {\frac {\pi }{4}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ceddb79275703a9fe7c7c64c64e6538c9d642de)
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
)
![{\displaystyle \arctan x=\sum _{n\geq 0}(-1)^{n}{x^{2n+1} \over {2n+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d9b9ad3b8512c3d470db5df66870c8c7d63e23a)
which is obtained integrating the geometric series ( absolutely convergent for
)
![{\displaystyle 1-x^{2}+x^{4}-x^{6}+x^{8}-\cdots ={\frac {1}{1+x^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52231ad15effa85b87352abebc8aa3e6a3ac0126)
termwise.