I was trying to help my friend find the limiting value of a recursive sequence, and tried something I feel is weakly justified and even plain wrong in a couple parts (I am not very familiar with recursive sequences to be honest, so I was making a stretch when trying to get the answer). The questions is:
Given $$U_{n+2} = 111 - \frac{1130}{U_{n+1}} + \frac{3000}{U_{n+1}U_n}$$ $$U_0 = 2, \space\space U_1 = -4$$ What value does this sequence approach as $n \to \infty$
my approach:
$$U_{n+2} = 111 - \frac{1130}{U_{n+1}} + \frac{3000}{U_{n+1}U_n}$$ $$U_{n+2} - U_{n+1} = \frac{U_{n+2} - U_{n+1}}{(n+2) - (n+1)} = 111 - \frac{1130}{U_{n+1}} + \frac{3000}{U_{n+1}U_n} - U_{n+1}$$
In the limit that $n \to \infty$, we have that $n+2 \approx n+1$, so: $$\frac{U_{n+2} - U_{n+1}}{(n+2) - (n+1)}$$ is equivalent to: $$\lim_{h \to 0} \frac{U(t + h) - U(t)}{h} = U'(t)$$ Where I am treating $t$ as a continuous variable representing $n$ in the large limit. Thus, we can say: $$U' = 111 - \frac{1130}{U} + \frac{3000}{U^2} - U$$ This has fixed point when $U' = 0$, or: $$U^3 - 111U^2 + 1130U - 3000 = 0$$ Which has solutions $U = 5, 6, 100$.
This is as far as I got. Saying $U' = f(U)$ will give you that $$f'(6) > 0, \space\space f'(5) < 0, \space\space f'(100) < 0$$
So that $U = 6$ is unstable and the other two are stable.
However, we know that the answer is $U = 6$, so this appears contrary to expectation. Moreover, I don't know any way to reason what the initial conditions of the differential equation $U' = f(U)$ are, because it depends on the large limit of the recursive map.
The fact that this method almost produced fruitful results is curious to me, I was just playing around. Can this idea be extended to something correct, or is it a lost cause?