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?

If I'm not getting it wrong, your series is
$$1-\frac 1 2 -\frac 13+\frac 14+\frac 15+\frac 16-\frac 17-\frac 18-\frac 19-\frac 1{10}+++++------\dots$$
So you have $1$ plus, $2$ minuses, $3$ pluses, $4$ minuses, and so on.
We can write your series as $a_0+a_1+a_2+a_3+\dots$ where
$$\begin{align} a_0&=(-1)^0 1 \\
 a_1&=(-1)^1\sum_{k=2}^{3}\frac 1 k\\
a_2&=(-1)^2\sum_{k=4}^{6}\frac 1 k\\
a_3&=(-1)^3\sum_{k=7}^{10}\frac 1 k\\
\cdots &=\cdots \\
a_n&=(-1)^n\sum_{k=T_n+1}^{T_{n+1}}\frac 1 k\end{align}$$
Where $T_n=\frac{n(n+1)}{2}$ so it goes $1,3,6,10,\dots$
Now, we know that $$\sum_{k=1}^n \frac 1 k =\log n+\gamma+\frac 1 {2n}+O(n^{-2})$$
Thus
$$\eqalign{
  & \sum\limits_{k = 1}^{{T_n}} {{1 \over k}}  = \log n + \log \left( {n + 1} \right) - \log 2 + \gamma  + O({n^{ - 2}})  \cr 
  & \sum\limits_{k = 1}^{{T_{n + 1}}} {{1 \over k}}  = \log \left( {n + 2} \right) + \log \left( {n + 1} \right) - \log 2 + \gamma  + O({n^{ - 2}}) \cr} $$
Whence, after simplification 
$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}}  = \log \left( {1 + {2 \over n}} \right) + O(n^{-2})$$
Recall that $$\log(1+x)=x+O(x^2)$$ so
$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}}  = \frac{2}{n} +O\left(\frac 1 {n^2}\right)$$
Since $$\sum (-1)^n \frac 1 n $$ and $$\sum n^{-2}$$ converge, so does your series.
