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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](http://en.wikipedia.org/wiki/Telescoping_series)2012-12-10

2 Answers 2

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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. thnx2012-12-10
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    @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.2012-12-10
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    I delete my answer, since you explained the prblem in detailed. :)2012-12-10
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    @Babak : Feel free to leave it there, you don't need to remove it just because I posted more stuff.2012-12-10
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    Patrick Da Silva. I know that there is nothing wrong in your answer but I can choose this answer as correct only after 10 minutes2012-12-10
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    @Barack Yes I agree Dont delete that2012-12-10
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Hint: Write the nth term as $\frac{1}{n}-\frac{1}{n+1}$.

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    Nice++++++++++++++++++12013-02-20
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    @amWhy: Thanks. It was a sooooo small hint. ;-)2013-02-20