Evaluate $$\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n+1)3^n}.$$
-A solution related to converting it into power series and then applying integration will be appreciated! I cant seem to express it as a power series.
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After some manipulation, it reduces to $-\log(1-z) = \sum_{n=1}^\infty \frac{z^n }{n}$ (antiderivative of the geometric series) – reuns Dec 10 '17 at 19:18
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Hint. Consider the power series expansion of $\arctan(x)$ at $0$: for $x\in(-1,1]$ $$\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)}=x+x\sum_{n=1}^{\infty}\frac{(-1)^n(x^2)^n}{(2n+1)}.$$ See Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?
Robert Z
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