Using $\tan^{-1}$ Show that $$ \pi = 2 \sqrt3 \sum_{n=0}^\infty  {(-1)^n\over(2n+1)3^n}$$
Really have no idea on this one guys. Its a practice question for my calc 2 final. Please help.
 A: HINTS:
First note that 
$$\arctan(x)=\int_0^x \frac1{1+t^2}\,dt$$
Second, represent $\frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
A: Hint:
$\;\dfrac\pi6=\arctan \biggl(\dfrac1{\sqrt 3}\biggr)$. Also remember that for $|x|<1$,
$$\arctan x=\sum_{n=0}^\infty(-1)^n \frac{x^{2n+1}}{2n+1}.$$
A: Recall that for $|t|<1$, 
$$\frac1t=\sum_{k\geq0}(1-t)^k$$
Hence we have that
$$\frac1{1+t^2}=\sum_{k\geq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$\int_0^x \frac{\mathrm dt}{1+t^2}=\arctan x$$
Doing the same on the LHS:
$$\int_0^x \sum_{k\geq0}(-1)^kt^{2k}\mathrm dt=\sum_{k\geq0}(-1)^k\int_0^x t^{2k}\mathrm dt=\sum_{k\geq0}(-1)^k\frac{x^{2k+1}}{2k+1}$$
Hence we have
$$\arctan x=\sum_{k\geq0}(-1)^k\frac{x^{2k+1}}{2k+1}$$
Now note that 
$$\arctan\bigg(\frac1{\sqrt{3}}\bigg)=\sum_{k\geq0}\frac{(-1)^k}{2k+1}\bigg(\frac1{\sqrt{3}}\bigg)^{2k+1}$$
Hence 
$$\frac\pi6=\sum_{k\geq0}\frac{(-1)^k}{(2k+1)3^{k+\frac12}}$$
$$\frac{\pi\sqrt{3}}6=\sum_{k\geq0}\frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$\frac\pi{2\sqrt{3}}=\sum_{k\geq0}\frac{(-1)^k}{(2k+1)3^k}$$
$$\pi=2\sqrt{3}\sum_{k\geq0}\frac{(-1)^k}{(2k+1)3^k}$$
QED
