Question

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

1/1.2 + 1/2.3 + …. + 1/ n.(n+1) v= 1- 1/n+1

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 - \frac1{1+1} = 1 - \frac12 = \frac12`

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 - (N+2)/((N+1)(N+2)) + \frac1{(N+1)(N+2)}`

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