Convergent/divergent series Is the following series divergent/convergent? 
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks 

We should be careful here. It's tempting to group the terms into natural subsums and argue the grouped series converges and leave it at that. But the convergence of a grouped series converges does not imply the original series converges in general. The obvious example being $1 + (-1) + 1 + (-1) +\cdots.$  Here's an example with the terms $\to 0$:
$$1-1 +1/2+1/2 -1/2 - 1/2 + 1/3+1/3 +1/3 - 1/3 -1/3-1/3 + \cdots.$$
One sufficient condition for the convergence of the grouped series to imply convergence of the orignal series is the following: Suppose the original series is $\sum_n a_n$ and we know the grouped series
$$(a_1 + \cdots + a_{n_1}) + (a_{n_1+1}+ \cdots +a_{n_2 }) + \cdots $$
converges. If
$$\lim_{k\to\infty} (|a_{n_k+1}|+ \cdots +|a_{n_{k+1}}|) = 0,$$
then the original series converges. I'll leave the proof to the reader as it's not too bad. (Briefly, the condition shows the partial sums of the original series differ from those of the grouped series by an amount $\to 0,$ giving the desired convergence.)
For the problem at hand, this is fairly easy to check.

Here's my own proof for this specific problem, where I don't appeal to the $\log $ function at all. For the $k$th grouping of positive and negative terms, let $M=(2k-2)(2k-1)/2, N = (2k-1)(2k)/2.$ Then the grouping is
$$\tag 1 \frac{1}{M+1} + \cdots + \frac{1}{M+2k-1} - \frac{1}{N+1}-\cdots - \frac{1}{N+2k}$$ $$ = \sum_{j=1}^{2k-1} \frac{N-M}{(M+j)(N+j)} - \frac{1}{N+2k}.$$
Note that $N-M = 2k-1.$ In absolute value then, making absolutely brazen estimates, $(1)$ is $\le (2k-1)^2/(M+1)^2 + 1/N.$ This is on the order of $1/k^2$ as $k\to \infty.$ This implies the grouped series converges absolutely, hence converges. How about the original series? It converges by the first part of this answer, because if we slap absolute values on each term in $(1),$ the sum is is no more than $(4k-1)/M+1 \to 0$ as $k\to 0.$ This finishes the proof.
