<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] Compare the absolute value of each homogeneously signed sum with the sum containing an equal number of copies of the first term.  For example, the all-negative terms from $-1/7$ to $-1/10$ are compared with four copies $-1/7$.  The absolute values of the sums then satisfy
$(1/7+1/8+1/8+1/9+1/10)<4/7$
In general the cluster sums have absolute values less than 
$n/({n(n-1)}/2+1)$
which goes to zero as $n$ increases without bound, so the series passes the test for convergence.
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