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How can I prove that $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? I noticed that in the sum, the denominator has terms that cancel out, but I'm not sure how to take advantage of that.

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marked as duplicate by Cameron Buie, Marvis, Asaf Karagila, Amzoti, 5pm Feb 20 '13 at 23:29

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Observe that $1=(i+1)-i$. Put that in your numerator and smoke it. – Rahul Feb 20 '13 at 22:53
up vote 7 down vote accepted

$$\dfrac1{i(i+1)} = \dfrac{i+1-i}{i(i+1)} = \dfrac{i+1}{i(i+1)} - \dfrac{i}{i(i+1)} = \dfrac1i - \dfrac1{i+1}$$ Hence, \begin{align} S_n & = \sum_{i=1}^n \dfrac1{i(i+1)} = \sum_{i=1}^n \left(\dfrac1i - \dfrac1{i+1} \right)\\ & = \left(1 - \dfrac12\right) + \left(\dfrac12 - \dfrac13\right) + \left(\dfrac13 - \dfrac14\right) + \cdots + \left(\dfrac1n - \dfrac1{n+1}\right)\\ & = 1 - \left(\dfrac12 - \dfrac12\right) - \left(\dfrac13 - \dfrac13\right) - \left(\dfrac14 - \dfrac14\right) - \cdots - \left(\dfrac1{n-1} - \dfrac1{n-1}\right) - \dfrac1{n+1}\\ & = 1 - \dfrac1{n+1} = \dfrac{n}{n+1} \end{align}

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Thanks, very clear. – mirai Feb 20 '13 at 23:11


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$$\sum_{i=1}^n\frac{1}{i(i+1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots +\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\ldots$$

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Seems three people type faster)

I'll be a bit more specific: this is called partial fraction expansion. Rewrite $\frac{1}{k(k+1)}$ as $\frac{A}{k} + \frac{B}{k+1} = \frac{A(k+1) + Bk}{k(k+1)}$ for some constants $A$ and $B$, then equate the coefficients: $$ A+B=0\\ A=1 $$ to get two fractions: $\frac{1}{k} - \frac{1}{k+1}$. With the summation you get a telescoping sum leaving only the first and the last terms: $1-\frac{1}{n+1}$

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