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?

To elaborate on Ross Millikan's answer and André Nicolas's comment, denote the $n$-th block sum (ignoring the sign) by $B_n$. Then
\begin{align}
B_n &= \sum_{i=1}^n \frac1{k_n+n},\\
k_n &= \frac{n(n-1)}{2},\\
k_{n+1} &= k_n+n = \frac{n(n+1)}{2} \le n^2.\tag{1}
\end{align}
Therefore
\begin{align}
B_n - B_{n+1}
&= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^{n+1} \frac1{k_{n+1}+i}\\
&= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^n \frac1{k_{n+1}+i} - \frac1{k_n+2n+1}\\
&= \sum_{i=1}^n \frac{n}{(k_n+i)(k_{n+1}+i)} - \frac1{k_n+2n+1}\ \text{ by } (1)\\
&\ge \sum_{i=1}^n \frac{n}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\
&= \frac{n^2}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\
&\ge \frac{1}{k_n+2n} - \frac1{k_n+2n+1}\ \text{ by } (1)\\
&> 0.
\end{align}
Hence the block sum is indeed monotonic decreasing and the alternating series test applies.
