Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?

I figured I'd take a different approach to it, by providing upper and lower bounds.
For an upper bound, we'll assume that all values in the positive sequences are the largest of them, and all values in the negative sequences are the smallest. That is, for instance...
$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} < \frac{3}{4}$
And
$\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}>\frac{4}{10} = \frac{2}{5}$
For the positive terms, we have that the sums are bounded above by $\frac{2n+1}{n(2n+1)+1} < \frac{1}{n}$, while the negative terms' sums are bounded below by $\frac{2}{2n+1} > \frac{1}{n+1}$. As such, we can say easily that the series cannot be larger than
$$
1+\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2
$$
Similarly, the positive sums are bounded below by $\frac{1}{n+1}$ and the negative sums are bounded above by $\frac{2n}{n(2n-1)+1} < \frac{1}{n-1}$. And so, the series cannot be smaller than
$$
  \frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n+1} - \frac{1}{n-1}\right) = -1
$$
As such, we know that the series must be less than 2 and greater than -1, so we know that the series does not diverge, and thus we only need to show that it does not fail to converge due to oscillation. As both the positive and negative sum terms behave as $1/n$ in the limit as $n \to \infty$ (thereby preventing cyclic behaviour), we know that the series must converge.
