Show that $\sum_1^\infty \frac{(-1)^{n-1}}{n} = \log 2 $ $$ \sum_1^\infty \frac{(-1)^{n-1}}{n} = \log 2 $$

I'm not sure how I can prove that.

I have tried changing the term to $-\cos(n\pi)/n$, but that doesn't help me much.  Any hint?  Thank you.
Recall the Taylor series for log
\begin{align*}
\log x=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}(x-1)^n.
\end{align*}
Plugging in $x=2$ solves your problem.
Here is an article that answers your question http://www.math.uconn.edu/~kconrad/blurbs/analysis/abelthm.pdf

It uses the power series for $\log(1+x)$ and Abel's theorem. If you are unfamiliar with this series, it comes from integrating the series $$\sum_{n=0}^{\infty} (-1)^nx^n=\frac{1}{1+x}.$$