SOLUTION: Compute the sum (2)/(1*2*3) + (2)/(2*3*4) + (2)/(3*4*5) + ...

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Question 1035833: Compute the sum
(2)/(1*2*3) + (2)/(2*3*4) + (2)/(3*4*5) + ...
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
2%2F%281%2A2%2A3%29+%2B+2%2F%282%2A3%2A4%29+%2B+2%2F%283%2A4%2A5%29++...
By partially decomposing the fraction, it is easy to see that
2%2F%28n%28n%2B1%29%28n%2B2%29%29+=+1%2Fn+-+2%2F%28n%2B1%29+%2B+1%2F%28n%2B2%29.
==> The partial sum up to the nth term is as follows:
1%2F1+-+2%2F2+%2B+1%2F3 +
1%2F2+-+2%2F3+%2B+1%2F4 +
1%2F3+-+2%2F4+%2B+1%2F5 +
1%2F4+-+2%2F5+%2B+1%2F6 +
1%2F5+-+2%2F6+%2B+1%2F7 +
..................
1%2F%28n-2%29+-+2%2F%28n-1%29+%2B+1%2Fn +
1%2F%28n-1%29+-+2%2Fn+%2B+1%2F%28n%2B1%29 +
1%2Fn+-+2%2F%28n%2B1%29+%2B+1%2F%28n%2B2%29.
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%2F2+-+1%2F%28n%2B1%29+%2B+1%2F%28n%2B2%29.
Therefore, letting n approach infinity, we obtain
2%2F%281%2A2%2A3%29+%2B+2%2F%282%2A3%2A4%29+%2B+2%2F%283%2A4%2A5%29++... = highlight%281%2F2%29.