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?

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 
$$
\arctan \dfrac1{\sqrt{3}}=\frac{\pi}6.
$$
