<s>[INST] Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing. 
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
 [/INST] So we have to deal with
$$ S=\sum_{k\geq 1}(-1)^{k+1} \sum_{n=\binom{k}{2}+1}^{\binom{k+1}{2}}\frac{1}{n} = \sum_{k\geq 2} (-1)^k A_k $$
and to prove convergence it is enough to show that $\{A_k\}_{k\geq 1}$ is decreasing (from some point on) and convergent to zero. The last claim is straightfoward to prove, since $A_k\geq 0$ but 
$$A_k\leq \frac{k}{\binom{k}{2}+1}\sim\frac{2}{k}. $$
The first claim can be easily proved by induction or a convexity argument:
$$ A_k = H_{\binom{k+1}{2}}-H_{\binom{k}{2}}, $$
hence $\{A_k\}$ is decreasing iff $k\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function. Since:
$$\begin{eqnarray*} H_n=\sum_{k=1}^{n}\frac{1}{k} &=& \sum_{k=1}^{n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)+\sum_{k=1}^{n}\log\frac{k+1}{k}\\&=&\gamma+\log(n+1)-\sum_{k>n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)\end{eqnarray*}$$
and $\left(\frac{1}{k}-\log\frac{k+1}{k}\right)$ is bounded between $\frac{1}{2k(k+1)}$ and $\frac{1}{2k^2}$, we have:
$$ H_{\binom{k+1}{2}}-H_{\binom{k}{2}} = \log\frac{k^2+k+2}{k^2-k+2}+O\left(\frac{1}{k^4}\right)$$
so $x\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function from some point on.
We may also notice that the value of our series depends on the integral:
$$ I =\int_{0}^{1}\frac{1+2\sum_{k\geq 1}(-1)^k x^{\binom{k}{2}}}{x-1}\,dx.
 $$
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