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.

You can apply Leibniz's alternating series test directly to the given series. First notice that, if $n$ is even, then:
\begin{align}
\log\left(1+(-1)^{n-1}\frac 1n\right)=\log\left(\frac{n-1}{n}\right)=-\log\left(\frac{n}{n-1}\right)=-\log\left(1+\frac{1}{n-1}\right)
\end{align}
So: $$\sum_{n=1}^\infty \log\left(1+(-1)^{n-1}\frac 1n\right) = \sum_{n=1}^\infty (-1)^{n-1}a_n$$ where \begin{align}
a_n=
\begin{cases}
\log(1+\frac{1}{n}) & \text{ if } & n \text{ is odd}\\
\log(1+\frac{1}{n-1}) & \text{ if } & n \text{ is even}
\end{cases}
\end{align}
Clearly $a_n$ is nonnegative and $a_{n+1}\leq a_n$ and $\lim_n a_n=0$ so by Leibniz's test the original series converges.
