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