# Rewrite function as Taylor series equal to natural logarithm of some value

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.

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$$-\log\left(\frac23\right)=\log\left(\frac32\right).$$