TITLE: Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$
QUESTION [6 upvotes]: Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$:
$$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$
The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.

REPLY [3 votes]: Alternatively, note that
$$S:=1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\ldots=\sum_{k=1}^\infty\,\frac{-\omega^{k}-\bar{\omega}^{k}}{k}\,,$$
where
$$\omega:=\exp\left(+\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{3}\text{i}}{2}\,,\text{ so }\bar{\omega}=\exp\left(-\frac{2\pi\text{i}}{3}\right)=\frac{-1-\sqrt{3}\text{i}}{2}\,.$$
Since the logarithmic power series $\ln(1-z)=\displaystyle- \sum_{k=1}^\infty\,\frac{z^k}{k}$ converges for all $z\in\mathbb{C}\setminus\{1\}$ such that $|z|\leq 1$, we get
$$S=\ln(1-\omega)+\ln(1-\bar{\omega})=\left(\frac{1}{2}\,\ln(3)-\text{i}\,\frac{\pi}{6}\right)+\left(\frac{1}{2}\,\ln(3)+\text{i}\,\frac{\pi}{6}\right)\,,$$
making $S=\ln(3)$.