Showing $\sum\limits_{n = 1}^∞\log(1+(-1)^{n-1}\frac{1}{n})$ is convergent Show that $$\sum_{n = 1}^\infty \log\left(1+(-1)^{n-1}\frac{1}{n}\right)$$ converges.
I want to say that the convergence/divergence of this series is equivalent to the convergence/divergence of $$\sum(-1)^{n-1}\frac{1}{n}.$$ Without the sign term I can show by L'hospital's rule that $$\lim_{n\to \infty}\frac{\log(1+1/n)}{1/n}=1.$$ But I don't know how to compare the given sries with $\sum(-1)^{n-1}/n$. Any suggestion is appreciated.

Just to add a note: Since $\ln(1+x)\leq x$ for all $x$, the convergence of 
$$\sum_{n=1}^{\infty} \ln(1+x_n)$$
is, in general, implied by that of
$$\sum_{n=1}^{\infty} x_n.$$
(In fact, if $x_n>0$, it is possible to say more; since the convergence of the first sum is euivalent to that of $\prod_{n=1}^{\infty} (1+x_n)$ by definition, this is equal to $1+x_1+x_2+\cdots+(x_1x_2+x_1x_3+x_2x_3+\cdots)+\cdots$ which is strictly greater than $\sum_{n=1}^{\infty} x_n$, so the two convergences are actually equivalent.)
