$\sum (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$ convergent but not absolutely convergent I need to prove that:
$$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$$
is convergent, but not absolutely convergent.
I tried the ratio test:
$$\frac{a_{n+1}}{a_n} = -\frac{\log\left(1+\frac{1}{n+1}\right)}{\log\left(1+\frac{1}{n}\right)} = -\log\left({\frac{1}{n+1}-\frac{1}{n}}\right)$$
I know that the thing inside the $\log$ converges to $1$, so $-\log$ converges to $0$? This is not right, I cannot conclude that this series is divergent.
Also, for the sequence without the $(-1)^{n+1}$ it would give $0$ too.

First i would show the absolute converge by the test for alternating series. First show that it is a alternating series (pretty obvious since your log only gives positive values). Then since $$ log(1+ \frac{1}{n} ) \rightarrow 0 \ , n \rightarrow \infty $$ it is convergent. To show the absolute i would remove $ (-1)^{n+1} $ since its only changing +-. Then i would use comparison test at compare to the divergent series $$ \frac{1}{n} $$
