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Toolkit 5

Telescoping series

11⋅2+12⋅3+⋯+1n(n+1)=(1−12)+⋯+(1n−1n+1)=1−1n+1\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)}=\left(1-\frac{1}{2}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}

Proof

For every positive integer kk,

1k(k+1)=1k−1k+1.\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

Thus,

11⋅2=1−12,\frac{1}{1\cdot 2} = 1 - \frac{1}{2},
12⋅3=12−13,\frac{1}{2\cdot 3} = \frac{1}{2} - \frac{1}{3},
⋮\vdots
1n(n+1)=1n−1n+1.\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.

Adding these identities, all intermediate terms cancel. Therefore,

11⋅2+12⋅3+⋯+1n(n+1)=1−1n+1.□\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}. \quad\square