Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$ 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.

This is a direct proof using $$\int_0^1 x^n dx= \frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.
$$\begin{align}
S&=\sum_{k=0}^\infty \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)\\
&=\sum_{k=0}^\infty \int_0^1 \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\
&=\int_0^1 \sum_{k=0}^\infty \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\
&=\int_0^1 \frac{1+x-2x^2}{1-x^3}dx\\
&=\int_0^1 \frac{1+2x}{1+x+x^2}dx\\
&=\log(1+x+x^2)|_{0}^1\\
&=\log(3)
\end{align}$$
Thus, similarly to
$$\log(1+x)=1-\frac{1}{2}x+\frac{1}{3}x^2-\frac{1}{4}x^3...$$
we have
$$\log(1+x+x^2)=1+\frac{1}{2}x-\frac{2}{3}x^2+\frac{1}{4}x^3+\frac{1}{5}x^4-\frac{2}{6}x^5+...$$
Moreover, the pattern generalizes. 
