2015/024) find the sum $\sum_{n=6}^{+\infty}\dfrac{2}{n(n+2)}$

this is telescopic sum
we have
$\dfrac{2}{n(n+2)}= \dfrac{1}{n}- \dfrac{1}{n+2}$

as n goes to inifinite the term  $\dfrac{1}{n}$ converges to zero and when we add we are left with

$\dfrac{1}{6}+\dfrac{1}{7}= \dfrac{13}{42}$