# What is the value of the following summation?

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?

• What about using $x=1/3$? – egreg Nov 27 '14 at 18:22
• @leo Consider up-voting and accepting answer if it helps! (by clicking on $\checkmark$) – Aditya Hase Nov 30 '14 at 3:19
• @Iuʇǝƃɹɐʇoɹ I like those special $\LaTeX$ symbols :D – Simply Beautiful Art Dec 10 '16 at 23:39

Hint: $\displaystyle \sum_{n=0}^\infty (-1)^{n+1}\dfrac{1}{9^n(2n+2)} = - 9\displaystyle \sum_{n=0}^\infty (-1)^n\dfrac{(\frac{1}{3})^{2n+2}}{2n+2}$