Answer
Use the Taylor series
$\pi=2\sqrt 3\sum_{k=0}^{\infty} \dfrac{(-1)^k }{3^k(2k+1)}$
Work Step by Step
In order to approximate $\pi$ we can use the function $f(x)=\tan^{-1} x$ and the Taylor series for $x=\dfrac{1}{\sqrt 3}$.
$\tan^{-1} x=\sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k+1}}{2k+1}$
$\tan^{-1} \dfrac{1}{\sqrt 3}=\sum_{k=0}^{\infty} \dfrac{(-1)^k \left(\dfrac{1}{\sqrt 3}\right)^{2k+1}}{2k+1}$
$\dfrac{\pi}{6}=\dfrac{1}{\sqrt 3}\sum_{k=0}^{\infty} \dfrac{(-1)^k \left(\dfrac{1}{3}\right)^k}{2k+1}$
$\pi=2\sqrt 3\sum_{k=0}^{\infty} \dfrac{(-1)^k }{3^k(2k+1)}$
