What is the sum of $\sum _{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$ I have the following series:
$$\sum _{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$$
I already know that it converges but I'm trying to get what does it sum up to. I've tried to find an alternative way of expressing it so that it becomes a geometric series but that got me nowhere. Any help is appreciated, thanks a lot.
 A: The Maclaurin series of the inverse tangent $\tan^{-1}$ is
$$\tan^{-1}(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
This holds for every $x\in[-1,1]$. Applying an index shift $n\rightarrow n-1$, we see that an equivalent representation is
$$\tan^{-1}(x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{2n-1}}{2n-1}$$
Given the obvious similarity between your series and this one, it makes sense to try and manipulate yours so it fits this form.
\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}} &= \sum_{n=1}^\infty (-1)^{n-1}\frac{\left(\frac{1}{\sqrt{3}}\right)^{2(n-1)}}{(2n-1)}\\
&= \sum_{n=1}^\infty (-1)^{n-1}\frac{\left(\frac{1}{\sqrt{3}}\right)^{2n-2}}{(2n-1)}\\
&= \left(\frac{1}{\sqrt{3}}\right)^{-1}\sum_{n=1}^\infty (-1)^{n-1}\frac{\left(\frac{1}{\sqrt{3}}\right)^{2n-1}}{(2n-1)}\\
&= \sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{align}
Using $\tan^{-1}\left(1/\sqrt{3}\right)=\pi/6$, we conclude that
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}=\frac{\pi\sqrt{3}}{6}$$
A: You may like this. Let
$$ f(x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{2n-1}}{2n-1}. $$
Then
$$ f'(x)=\sum_{n=1}^\infty (-1)^{n-1}x^{2n-2}=\sum_{n=1}^\infty (-x^2)^{n-1}=\frac{1}{1+x^2} $$
and hence
$$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}=\sqrt3\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1}\left(\frac1{\sqrt3}\right)^{2n-1}=\sqrt3\int_0^{1/\sqrt3}\frac{1}{1+x^2}dx=\frac{\sqrt3\pi}{6}. $$
