Müller series : solving a recursive sequence I came across this recursive sequence (often called the Müller sequence) defined as follows : 
$U_{n+2} = 111 - \frac{1130}{U_{n+1}}+\frac{3000}{U_{n+1}\times U_{n}} $, with $U_{0} = 2$ and $U_{1} = -4$ and I want to find the expression of $U_{n}$ in function of $n$.
I found in some forums that the closed expression is $\frac{4\times 5^{n+1}-3\times6^{n+1}}{4\times5^{n}-3\times6^{n}}$ but I have no clue how to show it mathematically.
The purpose of this sequence is to prove that mathematically, the limit is $6$, whilst when using Maple, Wolfram Alpha or any other computation tool, the resulting limit is $100$ (which I've already verified using multiple computation tools).
One approach I used is to suppose that $U_{n}$ converges to $l$ and then to take the resulting equation : $l^{3} - 111l^{2} + 1130l - 3000 = 0$, I found its roots : $l_{1} = 5$,  $l_{2} = 6$ and  $l_{3} = 100$. 
I don't know however how to show that, considering the initial conditions, only the valid answer is $6$.
I'd appreciate any new insights.
 A: Let $$V_n=\prod^n_{k=0}U_k\tag1,$$ meaning the empty produkt $V_{-1}$ equals $1$, so that we have $\displaystyle U_n=\frac{V_n}{V_{n-1}}$ even for $n=0$. Then, multiplying the given equation by $V_{n+1}$, we arrive at $$V_{n+2} = 111\,V_{n+1} - 1130\,V_n+3000\,V_{n-1},\tag2$$ with the initial conditions $V_{-1}=1, V_0=2, V_1=-8$.
As was pointed out by the OP, the roots of the characteristic equation are $5, 6, 100$, so the general solution of (2) is
$$V_n=A\cdot5^n+B\cdot6^n+C\cdot100^n.\tag3$$
It's easy to see that $$V_n=4\cdot 5^{n+1}-3\cdot6^{n+1}$$ satisfies the initial conditions, so it is our special solution, and from (1), we have
$$U_n=\frac{V_n}{V_{n-1}}=\frac{4\cdot 5^{n+1}-3\cdot6^{n+1}}{4\cdot 5^n-3\cdot6^n},$$ indeed. So $\displaystyle\lim_{n\to\infty}=6$ is obvious.
The catch is that it's practically impossible to recover that numerically. The component $C\cdot100^n$ in (3) is zero due to the special choice of the initial conditions. Any inevitable rounding error will introduce a spurious component with that behavior, and no matter how small the spurious $C$ is, the exponent $100^n$ will make it blow up. And as soon as that component is present, the limit will be $100$, of course. 
