sequence {u_n} of real numbers

[orl]

A sequence {u_n} of real numbers is given by u_1 and, for n \geq 1, by the recurrence relation 4*u_(n+1) = (64*u_n+15)^(1/3). Describe the behaviour of u_n for n->00.

IMO shortlist 1981

[Valentin Vornicu]

easy exercise.
consider the function: f(x)=64x³-64x-15. We can see that
f(x)=64x³-4x-60x-15=4x(4x+1)(4x-1)-15(4x+1)=(4x+1)(16x²-4x+15)=64(x+1/4)(x- (1+\sqrt 61)/8)(x-(1-\sqrt 61)/8) and using derivatives we have f'(x)=192x²-64 we see that the function is increasing on (-\infty , -1/ \sqrt 3) and (1/ \sqrt 3 , +\infty ) and decreasing on (-1/ \sqrt 3, 1/ \sqrt 3) and positive and negative on invervals given by the roots of f(x).

and now depending on the choice of u₁ we either have
case 1: u₁ > (1+\sqrt 61)/8 > 1> 1/ \sqrt 3 => by induction the sequence u_n is increasing, and it is not bounded, otherwise it would have to be convergent, but it is already larger than any of the possible three limits.

the other cases can be easily treated simillarly, and all of them will result in limits one of the three roots.