Compute $$\displaystyle\sum \limits_{n=0}^\infty (-1)^{n+1} \frac{1}{9^n(2n+2)}$$
I am given the fact that $$ \frac{1}{2}\ln(1+x^2) = \sum \limits_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{2n+2} $$ but I still don't have any clue how to calculate the sum for the first series. Any suggestions?
Hint: $\displaystyle \sum_{n=0}^\infty (-1)^{n+1}\dfrac{1}{9^n(2n+2)} = - 9\displaystyle \sum_{n=0}^\infty (-1)^n\dfrac{(\frac{1}{3})^{2n+2}}{2n+2}$