# Calculating the sum with Power Series

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.

• In essence, consider the powerseries with $x$ where $x$ has been replaced by $\pi/4$. – imranfat Sep 28 '16 at 3:14
• Ok done. $$-1\sum_{n=1}^\infty\frac{x^n}{(2n-1)}$$ But I'm still not seeing how to move on from this. – Carlos V Sep 28 '16 at 3:18
• the negative can go up front, we don't need that here – imranfat Sep 28 '16 at 3:18
• Hint: Start with geometric series, replace $x$ by $x^2$, then divide both sides by $x^2$ and integrate. See Ross' answer to finish it. (When integrating, don't forget to assume a constant) – imranfat Sep 28 '16 at 3:25

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