0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What about using $x=1/3$? $\endgroup$ – egreg Nov 27 '14 at 18:22
  • 2
    $\begingroup$ @leo Consider up-voting and accepting answer if it helps! (by clicking on $\checkmark$) $\endgroup$ – Aditya Hase Nov 30 '14 at 3:19
  • $\begingroup$ @Iuʇǝƃɹɐʇoɹ I like those special $\LaTeX$ symbols :D $\endgroup$ – Simply Beautiful Art Dec 10 '16 at 23:39
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.