<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] This answer uses similar ideas to those in Jack D'Aurizio's solution:
First we let $\displaystyle c_n=\frac{n^2-n+2}{2}$ and let $\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}$  for each n,
and we consider the related series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ obtained by grouping terms with the same sign,
so $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n=1-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\cdots$
1) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}\le\sum_{k=0}^{n-1}\frac{1}{c_n}=\frac{n}{c_n}=\frac{2n}{n^2-n+2}\to0,\;\;$ so $a_n\to 0$.
2) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}>\int_{c_n}^{c_n+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2n}{n^2-n+2}\right)$ for each n, and
$\;\;\;\;\displaystyle a_{n+1}=\sum_{k=0}^{n}\frac{1}{c_{n+1}+k}<\int_{c_{n+1}-1}^{c_{n+1}+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2}{n}\right)$ for each n.
$\hspace{.2 in}$Since $\displaystyle\frac{2n}{n^2-n+2}\ge\frac{2}{n}$ for $n\ge2$, $\;\;a_n > a_{n+1}$ for $n\ge 2$.
Therefore $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ converges by the Alternating Series Test.

Let $(T_n)$ be the sequence of partial sums of this series, and let $T_n\to T$.
If $(S_k)$ is the sequence of partial sums for the original series,
then for any value of $k$ let $n$ be the largest integer satisfying $n^2-n+2\le 2k$.
Then $\big|S_k-T\big|\le\max\{|T_n-T|, |T_{n+1}-T|\}$, so it follows that $S_k\to T$.
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