Evaluating $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$ 
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
 A: One can notice that 
$$\frac{1}{x^2}\int x^{2n}\mathrm dx=\frac{x^{2n-1}}{(2n+1)}$$
So:
$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\sum_{n=1}^{\infty} {(-1)^n }\int x^{2n}\mathrm dx=\frac{1}{x^2}\int \left(\sum_{n=1}^{\infty} {(-1)^n }x^{2n}\right)\mathrm dx$$
And (keeping in mind that $x\leq 1$)
$$\sum_{n=1}^{\infty} {(-1)^n }x^{2n}=-\frac{x^2}{x^2+1}$$
Then 
$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\int \left(-\frac{x^2}{x^2+1}\right)\mathrm dx=\frac{\tan^{-1}(x)-x}{x^2}$$
A: Since
$$
\tan^{-1}(x)=\sum_{k=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}
$$
we get
$$
\sum_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{2n+1}
=\frac{\tan^{-1}(x)-x}{x^2}
$$
For $x=\frac23$, the sum is approximately $-0.176994142017973$.

Without knowing the series for $\tan(x)$ first
It is not too difficult to notice that if we multiply the series by $x^2$ and differentiate, we get $$\sum_{k=1}^\infty(-1)^kx^{2k}=\frac1{1+x^2}-1$$ We get that by using the geometric series formula. Now we integrate and divide by $x^2$ to undo what was just done.
A: 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
