<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] $$
\underbrace{\vphantom{\frac11}+1}_{\text{length }1}
\underbrace{-\frac12-\frac13}_{\text{length }2}
\underbrace{+\frac14+\frac15+\frac16}_{\text{length }3}
\underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4}
\underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots
$$
The absolute values of the terms of the same-sign block of length $n$ are from
$$
\dfrac1{n(n-1)/2+1}\quad\text{to}\quad\dfrac1{n(n+1)/2}
$$
and the sum of the block must satisfy
$$
\frac2{n+1}=\frac{n}{n(n+1)/2}\le(-1)^{n-1}\text{sum}\le\frac{n}{n(n-1)/2+1}\lt\frac2{n-1}
$$
which tends to $0$.
The absolute value of the sum of the same-sign block of length $n$ and the same-sign block of length $n+1$ is at most
$$
\frac2{n-1}-\frac2{n+2}=\frac6{(n+2)(n-1)}
$$
Thus, the sum of pairs of blocks converge absolutely and the blocks converge to $0$. Thus, the full series converges.
</s>