If the question were $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=\log 2$
I would do it like this $\log(1+ z)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}z^n$ and since the series $\sum_n \frac{(-1)^{n+1}}{n}$ converges, it follows by Abel's Theorem that $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \lim_{r \to 1^-}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}r^n = \lim_{r \to 1^-} \log(1+r)=\log 2$
Would be correct?
Is it possible that there was a typo in the question? If not, the first term would go to infinity, right?
Anyway, I'm confused.
Thanks in advance for any help.