<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] 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.
</s>