<s>[INST] 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?
 [/INST] 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.
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