TITLE: Evaluating $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$
QUESTION [7 upvotes]: Calculate the summation $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$.

So I said:

Mark $x = \frac{2}{3}$. Therefore our summation is $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{x^{2n-1}}{(2n+1)}}$. 

But how do I exactly get rid of the $(-1)^n$? Also I notice it is a summation of the odd powers of $x$, how can I convert it to a full sum? (I know I should subtract from the full sum) but the signs of this summation is different than the signs of the full sum

REPLY [3 votes]: I'd start using a geometric series, let $|x|<1$ then:
$$\sum_{n=1}^\infty (-1)^n x^{2n}=-\frac{x^2}{1+x^2}=\frac{1}{1+x^2}-1.$$
Then by integrating with respect to $x$ and dividing by $x^2$ you get robjohn's result