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The left side is equivalent to
[tex]\displaystyle \sum_{k=1}^n \frac1{k(k+1)}[/tex]
When n = 1, we have on the left side
[tex]\displaystyle \sum_{k=1}^1 \frac1{k(k+1)} = \frac1{1\cdot2} = \frac12[/tex]
and on the right side,
[tex]1 - \dfrac1{1+1} = 1 - \dfrac12 = \dfrac12[/tex]
so this case holds.
Assume the equality holds for n = N, so that
[tex]\displaystyle \sum_{k=1}^N \frac1{k(k+1)} =1 - \frac1{N+1}[/tex]
We want to use this to establish equality for n = N + 1, so that
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}[/tex]
We have
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = \sum_{k=1}^N \frac1{k(k+1)} + \frac1{(N+1)(N+2)}[/tex]
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+1} + \frac1{(N+1)(N+2)}[/tex]
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+2}{(N+1)(N+2)} + \frac1{(N+1)(N+2)}[/tex]
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+1}{(N+1)(N+2)}[/tex]
[tex]\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}[/tex]
and this proves the claim.