Soit `(u_n)` la suite définie par `u_0 = 1 ` et `u_(n+1)= 1+1/(1+u_n) ` pour tout ` n in N `

1) Calculer ` u_1 , u_2 `

2) Montrer que ` forall n in N : 1 <= u_n <= 3/2 `

3) Montrer que `( forall n in N^(ast) ) : abs(u_(n+1) -u_n) <= 1/4abs(u_n -u_(n-1)) `

1) Calculer ` u_1 text{ et } u_2 `

on a `u_1 = u_(1+0)= 1+ 1/(1+u_0) `

` = 1+1/(1+1)= 3/2 `

`=> u_1 = 3/2 `

on a `u_2 = u_(1+1)= 1 +1/(1+u_1) = 1 +1/(1+3/2)`

` = 1+2/5 = 7/5 `

`=> u_2 = 7/5 `

2) Montrer que ` forall n in N : 1 <= u_n <= 3/2 `

Démonstration par récurrence

Initialisation

Pour ` n = 0 ` on a ` u_0 = 1 `

`=> 1 <= u_0 <= 3/2 `

Hérédité

Soit ` n in N ` on suppose que ` 1 <= u_n <= 3/2 ` montrons que ` 1 <= u_(n+1) <= 3/2 `

On a par hypothèse de récurrence ` 1 <= u_n <= 3/2 `

`=> 2 <= u_n +1 <= 3/2 + 1 = 5/2 `

`=> 2/5 <= 1/(u_n+1) <= 1/2`

`=> 2/5 + 1 <= 1+ 1/(u_n+1) <= 1/2 +1 `

`=> 1 < 2/5 + 1 <= 1+ 1/(u_n+1) <= 1/2 +1 `

`=> 1 <= u_(n+1) <= 3/2 `

Conclusion

Selon le principe de la récurrence on déduit que `( forall n in N ) : 1 <= u_n <= 3/2 `

3) Montrer que `( forall n in N^(ast) ) abs(u_(n+1) -u_n) <= 1/4abs(u_n -u_(n-1)) `

on a `u_(n+1) = 1+ 1/(1+u_n) ` et `u_n= 1+1/(1+u_(n-1))`

`=> u_(n+1) -u_n = 1+ 1/(1+u_n) - (1+1/(1+u_(n-1))) `

`=> u_(n+1) -u_n = 1/(1+u_n) -1/(1+u_(n-1)) `

`=> u_(n+1) -u_n = (1+u_(n-1) - 1-u_n)/((1+u_n)(1+u_(n-1)) `

`=> u_(n+1) -u_n = (u_(n-1) - u_n)/((1+u_n)(1+u_(n-1)) `

alors ` abs( u_(n+1) -u_n ) = (abs((u_(n-1) - u_n)))/((1+u_n)(1+u_(n-1)) ` car ` (1+u_n)(1+u_(n-1) ) > 0 `

Puisque ` forall n in N^(ast) : 1 <= u_n ` alors ` 2 xx2 <= (1+u_n)(1+u_(n-1) ) `

donc ` 1/((1+u_n)(1+u_(n-1) )) <= 1/4 `

`=> (abs(u_(n-1) - u_n))/((1+u_n)(1+u_(n-1))) <= 1/4 abs(u_(n-1) - u_n) `

`=> abs( u_(n+1) -u_n ) = (abs(u_(n-1) - u_n))/((1+u_n)(1+u_(n-1))) <= 1/4 abs(u_(n-1) - u_n) `

`( forall n in N^(ast) ) abs(u_(n+1) -u_n) <= 1/4abs(u_n -u_(n-1)) `