Question 1035833
{{{2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) }}}+...

By partially decomposing the fraction, it is easy to see that 

{{{2/(n(n+1)(n+2)) = 1/n - 2/(n+1) + 1/(n+2)}}}.

==> The partial sum up to the nth term is as follows:

{{{1/1 - 2/2 + 1/3}}} +
{{{1/2 - 2/3 + 1/4}}} +
{{{1/3 - 2/4 + 1/5}}} +
{{{1/4 - 2/5 + 1/6}}} +
{{{1/5 - 2/6 + 1/7}}} +
..................
{{{1/(n-2) - 2/(n-1) + 1/n}}} +
{{{1/(n-1) - 2/n + 1/(n+1)}}} +
{{{1/n - 2/(n+1) + 1/(n+2)}}}.

The effect of arranging the sum this way is, it becomes easy to see that diagonal "like" terms cancel each other, and what remains after the telescoping is the expression

{{{1/2 - 1/(n+1) + 1/(n+2)}}}.

Therefore, letting n approach infinity, we obtain

{{{2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) }}}+... = {{{highlight(1/2)}}}.