This should have been a fairly straightforward question: Show that $ln(\frac{2}{3}) = \sum_{n=1}^∞ \frac{-1^{n}}{2^n n}$
The Taylor Series expansion: $ln(1+x) = \sum_{n=1}^∞ \frac{-1^{n+1}}{n}x^n$ for $|x| < 1$ holds for $ln(\frac{2}{3})$, as the $x$ value would be $1 + x = \frac{2}{3}$, $x = - \frac{1}{3}$. Nevertheless, if you sub $x$ for $-\frac{1}{3}$, you get $ln(\frac{2}{3}) = \sum_{n=1}^∞ \frac{-1^{2n}}{-3^nn}$, which is not the same as the desired result.
Why is this the case?