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How do I rewrite $\sum_{n=1}^\infty\frac{\frac{1}{3}^n}{n}$ as a Taylor series of a function for some x in order to find the sum? I also know that the sum is the natural log of three-halves.

I have tried squaring the Taylor series of the natural logarithm of 1+x in order to get rid of the alternating part of the series, but this does not equal the correct sum.

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That sum is equal to $\sum_{n=1}^\infty\frac{x^n}n$, when $x=\frac13$. But$$x\in(-1,1)\implies\sum_{n=1}^\infty\frac{x^n}n=-\log(1-x)$$and so the sum of your series is$$-\log\left(\frac23\right)=\log\left(\frac32\right).$$

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