Olivia Vasquez

2022-02-25

That is equivalent to
$\pi =\sqrt{12}\sum _{n\ge 0}\frac{\left(-1{\right)}^{n}}{{3}^{n}\left(2n+1\right)}$

$\frac{\pi }{\sqrt{12}}=\sqrt{3}\mathrm{arctan}\left(\frac{1}{\sqrt{3}}\right)$
$=\sqrt{3}{\int }_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+{t}^{2}}dt$
$=\sqrt{3}{\int }_{0}^{\frac{1}{\sqrt{3}}}\sum _{n=0}^{\mathrm{\infty }}{\left(-{t}^{2}\right)}^{n}dt$
$=\sqrt{3}\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\int }_{0}^{\frac{1}{\sqrt{3}}}{t}^{2n}dt$
$=\sqrt{3}\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{\left[\frac{{t}^{2n+1}}{2n+1}\right]}_{0}^{\frac{1}{\sqrt{3}}}$
$=\sqrt{3}\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{2n+1}\left({\left(\frac{1}{\sqrt{3}}\right)}^{2n+1}-0\right)$
$=\frac{\sqrt{3}}{\sqrt{3}}\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{\left(\sqrt{3}\right)}^{2n}\left(2n+1\right)}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{3}^{n}\left(2n+1\right)}$
So
$\pi =\sqrt{12}\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{3}^{n}\left(2n+1\right)}$

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