Recursive sequence problem... Consider the recursive sequence
\begin{equation*}
\begin{split}
a_{n+1} = \frac{5}{6-a_n} \quad \textit{with} \quad a_1 = 4.
\end{split}
\end{equation*}
Prove that the sequence $(a_n)$ converges and find its limit, by working out the following steps.
1. First assume that the limit $L = \lim_{n\to \infty} a_n$ exists and find its possible values.
Let $L = \lim_{n\to\infty}a_n$. Then we get
\begin{equation*}
\begin{split}
L &= \lim_{n\to\infty}a_{n+1} \\
&= \lim_{n\to\infty} \frac{5}{6-a_n} \\
&= \frac{5}{6-L}.
\end{split}
\end{equation*}
So we have $L^2-6L+5 = 0 \Longleftrightarrow (L-1)(L-5) = 0$. So the possible values of $L$ are $L = 1$ or $L = 5$.
2. Starting with the initial value $a_1 = 4$, write down the first five entries in the sequence $(a_n)$. Can you see any pattern?
We have
\begin{equation*}
\begin{split}
a_2 &= \frac{5}{6-a_1} = \frac{5}{6-4} = \frac{5}{2} = 2.5 \\
a_3 &= \frac{5}{6-a_2} = \frac{5}{6-5/2} = \frac{10}{7} \approx 1.42857 \\
a_4 &= \frac{5}{6-a_3} = \frac{5}{6-10/7} = \frac{35}{32} = 1.09375 \\
a_5 &= \frac{5}{6-a_4} = \frac{5}{6-35/32} = \frac{160}{157} \approx 1.01091 \\
a_6 &= \frac{5}{6-a_5} = \frac{5}{6-160/157} = \frac{785}{782} \approx 1.00384.
\end{split}
\end{equation*}
3. Find the real valued function $f(x)$ defining the sequence, i.e. $a_{n+1} = f(a_n)$.
This is the question I'm having trouble with. Please help!
 A: Hints. For $3.$ $f(x)=\frac{5}{6-x}$ with $f'(x)=\frac{5}{(6-x)^2}>0$, i.e. the function is ascending. Now $a_2=f(a_1)=\frac{5}{2}<4=a_1$. Then $f(a_2)\leq f(a_1) \Rightarrow a_3\leq a_2$ and by induction
$$a_n\leq a_{n-1} \Rightarrow f(a_n)\leq f(a_{n-1}) \Rightarrow a_{n+1}\leq a_n$$
or the sequence is descending. 
Let's show that it is also bounded. From $$x\in[1,5] \Rightarrow 1\leq x \leq 5 \Rightarrow 5\geq 6-x \geq 1 \Rightarrow  1\leq \frac{5}{6-x}\leq 5$$
or
$$x\in[1,5] \Rightarrow f(x)\in[1,5]$$
Now, $a_1\in[1,5] \Rightarrow a_2=f(a_1)\in[1,5]$ and, again by induction, $a_n\in[1,5]$. So, the sequence is bounded in monotone.
A: The recuurence relation is $$u_{n+1}(u_n-6)=-5~~~(1)$$
Let $$u_{n}-6=\frac{v_{n-1}}{v_{n-2}}~ in ~(1).$$
We get $$v_n+6 u_{n-1}+5 v_{n-2}~~~(2)$$
Let $v_n=x^n$ in (2), we get $$x^2+6x+5=0 \Rightarrow
x_1=-5,x_2=-1$$; then $$v_n=C_{1} (-5)^n+ C_2(-1)^n~~~(3)$$
$$u_n=\frac{C_1 (-5)^{n-1}+ C_2(-1)^{n-1}}{C_1 (-5)^{n-2}+ C_2(-1)^{n-2}}+6.$$
The single unknown $D=C_1/ C_2$ can be determined by the initial values of $u_n,~$ namely $u_1=4$. Here $D$ coes out to be $5/3$, then
$$U_n=6+\frac{5(-5)^{n-1}+3(-1)^{n-1}}{5(-5)^{n-2} +3 (-1)^{n-2}}.$$
ehere it follows that $\lim_{n \rightarrow  \infty} u_n=1.$
