Question

Prouvez par récurrence que quel que soit n EN\{0}, on a

Answer 1

The left side is equivalent to

$\displaystyle \sum_{k=1}^n \frac1{k(k+1)}$

When n = 1, we have on the left side

$\displaystyle \sum_{k=1}^1 \frac1{k(k+1)} = \frac1{1\cdot2} = \frac12$

and on the right side,

$1 - \dfrac1{1+1} = 1 - \dfrac12 = \dfrac12$

so this case holds.

Assume the equality holds for n = N, so that

$\displaystyle \sum_{k=1}^N \frac1{k(k+1)} =1 - \frac1{N+1}$

We want to use this to establish equality for n = N + 1, so that

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

We have

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = \sum_{k=1}^N \frac1{k(k+1)} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+1} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+2}{(N+1)(N+2)} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+1}{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

and this proves the claim.