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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.

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  • $\begingroup$ In essence, consider the powerseries with $x$ where $x$ has been replaced by $\pi/4$. $\endgroup$ – imranfat Sep 28 '16 at 3:14
  • $\begingroup$ Ok done. $$-1\sum_{n=1}^\infty\frac{x^n}{(2n-1)}$$ But I'm still not seeing how to move on from this. $\endgroup$ – Carlos V Sep 28 '16 at 3:18
  • $\begingroup$ the negative can go up front, we don't need that here $\endgroup$ – imranfat Sep 28 '16 at 3:18
  • $\begingroup$ 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) $\endgroup$ – imranfat Sep 28 '16 at 3:25
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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.

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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}

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