How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
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Your last term on the left should have * instead of / in the denominator and the right side should have a / in it. – Ross Millikan Feb 11 '13 at 19:26
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1Please don't edit out your questions. If you feel that your question was answered you can accept an answer by clicking the check mark below the vote count. You can only accept one answer, but you can vote them all up or down if you wish to thank the users. – Asaf Karagila Feb 11 '13 at 20:37
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Very similar question: Sum of special series: $1/(1\cdot2) + 1/(2\cdot3 )+ 1/(3\cdot4)+\cdots$ – Martin Sleziak Feb 20 '13 at 13:13
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Several nice answers have been provided, which show how you can find out what the sum is. But if you were told in your assignment what the sum is supposed to be, you can prove this by induction. In the inductive step you only need to verify that $\frac{n-1}n+\frac1{n(n+1)}=\frac{n}{n+1}$. – Martin Sleziak Feb 20 '13 at 13:16
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User defacing one's questions, probably to cover their tracks. Flagged. – Did May 06 '13 at 19:39
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Please do not deface your questions. – robjohn May 06 '13 at 20:12
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$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\ldots+\frac{1}{n(n+1)}$
$=(1-\frac12)+(\frac12-\frac13)+\ldots+(\frac{1}{n}-\frac{1}{n+1})$
$=1-\frac{1}{n+1}$
$=\frac{n}{n+1}$
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The archetypal example of a http://en.wikipedia.org/wiki/Telescoping_series – Andreas Caranti Feb 20 '13 at 13:16
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$$\sum_{i=1}^{n}\frac{1}{i(i+1)}=\sum_{i=1}^{n}\Big(\frac{1}{i}-\frac{1}{i+1}\Big)=$$ $$=1+\sum_{i=2}^{n}\frac{1}{i}-\Big(\sum_{j=2}^{n}\frac{1}{j}+\frac{1}{n+1}\Big)=1-\frac{1}{n+1}$$
Adi Dani
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