<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] We can see that the general term of this series is
$$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$
and we have 
$$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$
so 
\begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+1)(n+2)}-\frac{1}{n(n+1)}+O(\frac{1}{n^2})\\
&=\frac{2}{n}+O(\frac{1}{n^2})\end{align}
hence we have
$$a_n=\frac{2(-1)^n}{n}+O(\frac{1}{n^2})$$
which  allows us to conclude the convergence of the series since it's sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series.
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