Calculating the sum with Power Series

Question

We're stuck on trying to evaluate the sum of this function using the power series, if you guys could provide some tips to continue through the process thank you. $$\sum_{n=1}^\infty\frac{-\pi^n}{4^n(2n-1)}$$ We tried moving the 4 up and factoring giving us $$\sum_{n=1}^\infty\frac{\frac{-\pi}{4}^n}{(2n-1)}$$ But still nothing to move on from, any help would be appreciated.

Answer 1

Can you sum $\sum_{n=1}^\infty x^{2n-2}$? Integrate it term by term and you get $\sum_{n=1}^\infty \frac {x^{2n-1}}{2n-1}$ Now multiply by $x$, set $x=\frac {\sqrt \pi}2$, and negate it.

Answer 2

For $\lvert x \rvert \lt 1,$ we have $$\tanh^{-1}(x)=\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}.$$

So you can write your sum as

\begin{align} \sum_{n=1}^\infty\frac{-\pi^n}{4^n(2n-1)}&=-\sum_{n=1}^\infty\frac{(\sqrt{\pi}/2)^{2n}}{2n-1} \\&=-\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{(\sqrt{\pi}/2)^{2n-1}}{2n-1} \\&=-\frac{\sqrt{\pi}}{2} \tanh^{-1}\left(\frac{\sqrt{\pi}}{2}\right). \end{align}