How to prove $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? [duplicate]

summation induction

Solution 1:

$$\dfrac1{i(i+1)} = \dfrac{i+1-i}{i(i+1)} = \dfrac{i+1}{i(i+1)} - \dfrac{i}{i(i+1)} = \dfrac1i - \dfrac1{i+1}$$ 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}

Solution 2:

$$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}.$$

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