1. ## series

$\sum_{n=1}^\infty \frac{6}{n(n+2)}$

$\frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$

$S_n = (6-\frac{6}{3}) + (\frac{6}{2}-\frac{6}{4}) + (\frac{6}{3}-\frac{6}{5})+$... $+ (\frac{6}{n} - \frac{6}{n+2})$

not sure what to do next

2. Originally Posted by viet

$\frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$

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The above is not true.

3. Originally Posted by viet
$\sum_{n=1}^\infty \frac{6}{n(n+2)}$

$\frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$
First that is wrong.

It should be,
$\frac{6}{n(n+2)} = \frac{3}{n} - \frac{3}{n+2}$

Now, here is how I explain it to people, look at the partial sums.

$S_1= \frac{3}{1} - \frac{3}{3}$

$S_2 = \frac{3}{1} - \frac{3}{3} + \frac{3}{2} - \frac{3}{4}$

$S_3 = \frac{3}{1} - \frac{3}{3}+\frac{3}{2} - \frac{3}{4} + \frac{3}{3} - \frac{3}{5}= \frac{3}{1} + \frac{3}{2} - \frac{3}{4} - \frac{3}{5}$

$S_4 = \frac{3}{1}+\frac{3}{2} - \frac{3}{4} - \frac{3}{5} + \frac{3}{4} - \frac{3}{6} = \frac{3}{1}+\frac{3}{2} - \frac{3}{5} - \frac{3}{6}$

It seems the pattern is,
$S_n = \frac{3}{1} + \frac{3}{2} - \frac{3}{n+1} - \frac{3}{n+2}$

Now just see what happens as $n\to \mbox{Hacker}$.