Let $\sum_{n=0}^\infty \frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge? 
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?

Lemma: Suppose $a_n\to 0$ and $\sum_{n=1}^{\infty}(a_{2n-1}+a_{2n})$ converges. Then $\sum_{n=1}^{\infty} a_n$ converges. 
Proof: Let $S_n$ denote the $n$th partial sum of $\sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}\to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_n\to 0.$ We also have
$$a_{2n-1}+a_{2n} =\frac{1}{3(2n-1) -6}-\frac{1}{3(2n)+1) -6} = \frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $\sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $\sum a_n$ converges.
