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The question is:

Find the sum of the series $$ 1/(1\cdot 2) + 1/(2\cdot3)+ 1/(3\cdot4)+\cdots$$

I tried to solve the answer and got the $n$-th term as $1/n(n+1)$. Then I tried to calculate $\sum 1/(n^2+n)$. Can you help me?

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The series $\sum\limits_{n=1}^{\infty}{\dfrac{1}{n(n+1)}}$ is Telescoping series –  M. Strochyk Dec 10 '12 at 8:52

2 Answers 2

up vote 8 down vote accepted

You have $$ \sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left( \frac 1i - \frac 1{i+1} \right) \\ = \sum_{i=1}^n \frac 1i - \sum_{i=1}^n \frac 1{i+1} \\ = \sum_{i=1}^n \frac 1i - \sum_{i=2}^{n+1} \frac 1i \\ = 1 - \frac 1{n+1}. $$ (we say that the sum telescopes). Therefore, if you let $n \to \infty$, the series converges to $$ \lim_{n \to \infty} 1 - \frac 1{n+1} = 1. $$ Hope that helps,

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n/n+1 is the answer. thnx –  chndn Dec 10 '12 at 8:50
    
@chndn : $1 - \frac 1{n+1} = n/(n+1)$, so there's nothing wrong there. I just thought that the limit in this form was easier to compute. –  Patrick Da Silva Dec 10 '12 at 8:51
    
I delete my answer, since you explained the prblem in detailed. :) –  Babak S. Dec 10 '12 at 8:51
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@Babak : Feel free to leave it there, you don't need to remove it just because I posted more stuff. –  Patrick Da Silva Dec 10 '12 at 8:52
    
Patrick Da Silva. I know that there is nothing wrong in your answer but I can choose this answer as correct only after 10 minutes –  chndn Dec 10 '12 at 8:53

Hint: Write the nth term as $\frac{1}{n}-\frac{1}{n+1}$.

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Nice++++++++++++++++++1 –  amWhy Feb 20 '13 at 16:07
    
@amWhy: Thanks. It was a sooooo small hint. ;-) –  Babak S. Feb 20 '13 at 16:21

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