Let $\displaystyle \sum_{n=0}^\infty \frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
I'm not completely sure that my calculation is correct, check it please.
$$\begin{align}\sum_{n\ge 0} \frac{(-1)^{n+1}}{3 n+6 (-1)^n}&=-\frac13\sum_{n\ge 0} \frac{(-1)^{n}}{n+2 (-1)^n}\cdot\frac{(-1)^n}{(-1)^n}\\&=-\frac13\sum_{n\ge 0} \frac1{n(-1)^n+2}\\&=-\frac13\lim_{m\to\infty}\left(\sum_{n=0\\ 2\mid n}^m\frac1{n+2}+\sum_{n=0\\ 2\nmid n}^m\frac1{-n+2}\right)\\&=-\frac13\lim_{m\to\infty}\left(\sum_{n=2\\ 2\mid n}^{m+2}\frac1{n}-\sum_{n=-2\\ 2\nmid n}^{m-2}\frac1{n}\right)\end{align}$$
If $m$ is even then
$$\begin{align}\lim_{m\to\infty}\left(\sum_{n=2\\ 2\mid n}^{m+2}\frac1{n}-\sum_{n=-2\\ 2\nmid n}^{m-2}\frac1{n}\right)&=\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m-2}\frac{(-1)^{n}}n\right)+\frac1m+\frac1{m+2}+1\right)\\&=-\log (2)+1\end{align}$$
If $m$ is odd then
$$\lim_{m\to\infty}\left(\sum_{n=2\\ 2\mid n}^{m+2}\frac1{n}-\sum_{n=-2\\ 2\nmid n}^{m-2}\frac1{n}\right)=\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m-1}\frac{(-1)^{n}}n\right)+\frac1{m+1}+1\right)=-\log (2)+1$$
Then finally
$$\bbox[2pt,border:yellow solid 2px]{\sum_{n=0}^\infty\frac{(-1)^{n+1}}{3 n+6 (-1)^n}=\frac{\log(2)-1}{3}}$$
Are my calculations correct?
One may set $$ S_N=\sum_{n=0}^N \frac{(-1)^{n+1}}{3 n+6 (-1)^n},\quad N\ge0, \tag1 $$ giving $$ \begin{align} S_{2N}&=\sum_{p=0}^N \frac{-1}{6p+6}+\sum_{p=1}^{N} \frac{1}{6p-9} \\\\S_{2N}&=-\frac16+\frac13\log 2+\psi\Big(N-\frac12 \Big)-\psi\left(N+2 \right) \tag2 \end{align} $$ and, as $N \to \infty$, giving $$ \lim_{N \to \infty}S_{2N}=-\frac13+\frac13\log 2, \tag3 $$ where $\psi$ denotes the standard digamma function.
Similarly,$$ \begin{align} S_{2N+1}&=\sum_{p=0}^N \frac{-1}{6p+6}+\sum_{p=0}^{N} \frac{1}{6p-3} \\\\&S_{2N+1}=-\frac13+\frac13\log 2+\psi\Big(N+\frac12 \Big)-\psi\left(N+2 \right) \end{align} $$ and, as $N \to \infty$, it gives $$ \lim_{N \to \infty}S_{2N+1}=-\frac13+\frac13\log 2. $$
We thus have
in agreement with your calculations.