AoPS - Problem

Consider the function

$f : \mathbb{R} \to \mathbb{R}$ given by

$f(x) = \sqrt[3]{x + \dfrac {15} {64}}$ . It has one real root at

$r = -\dfrac {15} {64}$ , it is increasing, convex on

$(-\infty,r]$ and concave on

$[r, +\infty)$ . The equation

$f(x) = x$ has three real roots at

$\rho_1 = \dfrac {1} {8}(1-\sqrt{61})$ ,

$\rho_2 = -\dfrac {1} {4}$ and

$\rho_3 = \dfrac {1} {8}(1+\sqrt{61})$ . On the other hand we have

$u_{n+1} = f(u_n)$ . The behaviour of

$u_n$ as

$n \to \infty$ depends on the starting term

$u_1$ , according to well-known techniques that resume to just considering the graph of

$f(x)$ and its intersection with the main bisector

$y=x$ , and will be described by:

$\bullet$

$\lim_{n \to \infty} u_n = \rho_1$ for

$x \in (-\infty, \rho_2)$ ;

$\bullet$

$\lim_{n \to \infty} u_n = \rho_2$ for

$x = \rho_2$ ;

$\bullet$

$\lim_{n \to \infty} u_n = \rho_3$ for

$x \in (\rho_2, +\infty)$ .