Use the power series for $\arctan(x)$ to derive an expression for $\pi$

Question

Given that the power series for $\arctan(x)$ is given by $$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$ for $x\in [-1,1]$ how would you show that $$\pi=2\sqrt3\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^n}?$$

I've tried by plugging in $x=1$ to get $$\frac{\pi}{4}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}$$ but other then plugging in values I have no clue how to reach the conclusion

Answer 1

$\tan \frac {\pi}{6} =\frac {1}{\sqrt 3}\\ \pi = 6\arctan \frac {1}{\sqrt 3}\\ \arctan x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n+1}}{2n+1}\\ \pi = 6\arctan \frac {1}{\sqrt 3} = 6\sum \frac {(-1)^n((\frac 13)^{\frac 12})^{2n+1}}{2n+1}\\ \pi = 6\frac {\sqrt 3}{3}\sum \frac {(-1)^n(\frac 13)^{n}}{2n+1}\\ \pi = 2\sqrt 3\sum \frac {(-1)^n}{3^n(2n+1)}\\ $

Answer 2

\begin{align*} \tan^{-1}(x/\sqrt{3})=\sum_{n=0}^{\infty}(-1)^{n}\dfrac{1}{2n+1}\left(\dfrac{x}{\sqrt{3}}\right)^{2n+1}=\dfrac{1}{\sqrt{3}}\sum_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n+1}}{(2n+1)3^{n}}, \end{align*} taking $x=1$, then \begin{align*} \pi=\dfrac{6}{\sqrt{3}}\sum_{n=0}^{\infty}(-1)^{n}\dfrac{1}{(2n+1)3^{n}}. \end{align*}