How do you prove $\pi =\sqrt{12}\sum_{n\ge 0}\frac{(-1)^n}{3^n(2n+1)}$?

In fact $\frac{\pi}{6}=\arctan\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\sum_{n\ge0}\frac{(-1)^n}{3^n(2n+1)}$, so $\sqrt{12}\sum_{n\ge0}\frac{(-1)^n}{3^n(2n+1)}=\sqrt{12}\sqrt{3}\frac{\pi}{6}=\pi$.

\begin{align*} \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{3^n(2n+1)}&=\sqrt{3}\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(\frac{1}{\sqrt{3}})^{2n+1}}{2n+1}\\ &=\sqrt{3}\arctan(\frac{1}{\sqrt{3}})\\ &=\sqrt{3}\frac{\pi}{6}\\ &=\frac{\pi}{\sqrt{12}} \end{align*}