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?
 A: 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.
A: It is correct. From
$$
\sum _{n=1}^{\infty } \frac{(-1)^n x^{2 n+1}}{2 n+1}=-x+\arctan x, \quad |x|<1,
$$ you get, with $x=\dfrac1{\sqrt{3}}$:
$$
\sum _{n=1}^{\infty } \frac{(-1)^n }{(2n+1)3^n}=\sqrt{3}\left(-\dfrac1{\sqrt{3}}+\arctan \dfrac1{\sqrt{3}}\right)=-1+\frac{\sqrt{3}\pi}6,
$$ using 
$$
\arctan \dfrac1{\sqrt{3}}=\frac{\pi}6.
$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}3^{n}}} & =
\sum_{n = 1}^{\infty}\pars{-\,{1 \over 3}}^{n} \int_{0}^{1}x^{2n}\,\dd x =
\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-\,{x^{2} \over 3}}^{n}\,\dd x =
\int_{0}^{1}{-x^{2}/3 \over 1 - \pars{-x^{2}/3}}\,\dd x
\\[5mm] & =
-\pars{\int_{0}^{1}\,\dd x
- \root{3}\int_{0}^{\root{3}/3}{\dd x \over x^{2} + 1}} =
- 1 + \root{3}\arctan\pars{\root{3} \over 3}
\\[5mm] & = \color{#f00}{{\root{3} \over 6}\,\pi - 1} \approx -0.0931
\end{align}
