<s>[INST] Trying to find an explicit sum of an infinite series: $\sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }$ I have the following series 
$$\sum_{n=1}^{\infty}\frac{(-1)^n }{(2n+1)3^n}$$
I am trying to find an explicit sum. I know this looks like $\arctan x = \sum_{n=0}^{\infty} \frac{ (-1)^n x^{2n+1}}{2n+1} $. I do the following 
$$ \sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } - 1 =\sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1$$
Since $(1/3)^n = (\sqrt{1/3})^2n = \sqrt{3} (\sqrt{1/3})^{2n+1}$. Thus, the sum is 
$$ \sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1)  } - 1 = \sqrt{3} \arctan(1/\sqrt{3}) = 1 = \boxed{\frac{ \sqrt{3} \pi }{6} - 1 }$$
Is this a correct solution? Do you guys a differenti method?
 [/INST] As $\log\dfrac{1+x}{1-x}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{2r+1}$
$$ \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }=\dfrac1{i/\sqrt3}\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}$$
Now $$2\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}=\log\dfrac{1+i/\sqrt3}{1-i/\sqrt3}=\log\dfrac{\sqrt3+i}{\sqrt3-i}=\log\dfrac{e^{i\pi/6}}{e^{-i\pi/6}}=\log(e^{i\pi/3})=\dfrac{i\pi}3$$
Considering principal values.
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