# Infinite power series sum $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$

Using theorems about differentiation or integration of power series calculate infinite sum of

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$$

The answer should equal to $$\frac{\pi}{2\sqrt3}$$.

I tried using $$f(x) = \arctan(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}x^{2n+1}$$ with $$x=\frac{1}{3}$$ but that fails, since we have $$3^n$$ and not $$3^{2n+1}$$ in the exponent.

• math.stackexchange.com/questions/1238780/… May 5 '19 at 9:40
• Try a different $x$. May 5 '19 at 9:42
• $3^n=\frac{1}{\sqrt{3}}\cdot \sqrt{3}^{2n+1}$
– Sil
May 5 '19 at 9:46

Factor an $$x$$ out entirely out of the sum, so you’re left with $$x^{2n}$$, then take $$x=\frac1{\sqrt{3}}$$.
Since$$\arctan(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1},$$you have$$\frac\pi6=\arctan\left(\frac1{\sqrt3}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n\sqrt3}$$and therefore$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n}=\frac{\pi\sqrt3}6=\frac\pi{2\sqrt3}.$$