<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] Michael's grouping answer works in general.  Note that the group with $n$ terms is of the form
$${1\over k+1}+{1\over k+2}+\cdots+{1\over k+n}$$
so the next group is
$${1\over k+n+1}+{1\over k+n+2}+\cdots{1\over k+2n}+{1\over k+2n+1}$$
It's easy to see that
$${1\over k+j}-{1\over k+n+j}={n\over(k+j)(k+n+j)}\gt{n\over(k+n)(k+2n+1)}$$
for each $1\le j\le n$, and so to show that the group with $n$ terms is bigger than the next group, it suffices to show
$${n^2\over(k+n)(k+2n+1)}\ge{1\over k+2n+1}$$
which is to say,
$$n^2\ge k+n$$
But this is easy:  The number that ends the group with $n$ terms, $k+n$, is the triangular number $n(n+1)\over2$.  And $n^2\ge{n(n+1)\over2}$ for all $n\ge1$.
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