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?
 A: 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

$$
\sum_{n=0}^\infty \frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-\frac13+\frac13\log 2
$$

in agreement with your calculations.
A: 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.
