Find an "upper bound" for a given sequence. Let $a_1=5$ and let $$a_{n+1}=\frac{a_n^2}{a_n^2-4a_n+6}$$
Find the biggest integer $m$ not bigger than $a_{2018}$, that is $m\leq a_{2018}$.
My go:
Apparently the limit must satisfy $$l=\frac{l^2}{l^2-4l+6}\Leftrightarrow l=0\vee l=3\vee l=2$$ Computing first few terms i see that $a_n\to 3$ as $n\to\infty$. The sequence seems to converge to 3, so i tried to show that $\forall n\geq 2,a_n\leq3$ proceeding by induction, we find that $a_1=5,a_2=25/11\approx2,27\leq3$. Now let $a_n\leq 3\Rightarrow a_{n+1}=\frac{a_n^2}{a_n^2-4a_n+6}\leq\frac{9}{a_n^2-4a_n+6}$ but here I'm stuck again, no idea what to do with the denominator. Any help appreciated.
 A: Write $b_n=a_n-3$.
Then
$$b_{n+1}=-3+b_{n+1}=\frac{-2a_n^2+12a_n-18}{a_n^2-4a_n+6}
=\frac{-2b_n^2}{b_n^2-2b_n+3}.$$
If $b_n$ is near $0$, then $b_{n+1}$ is negative, and approximately
$-2b_2^2/3$, and so is even nearer zero. Therefore if the sequence $(a_n)$
comes anywhere near $3$, it will converge rapidly to $3$, from below.
It does. Therefore the integer part of $a_{2018}$ will be $2$.
A: Suppose we compute the fixed points of the iteration $x\mapsto\frac{x^2}{x^2-4x+6}$; those fixed points are 0, 2 and 3. Now analyse the stability of those fixed points; a fixed point $x_0$ of $x\mapsto f(x)$ is stable (attractive) if $|f'(x_0)|<1$.
The numerator of the derivative of the given map is
$$2x(x^2-4x+6)-x^2(2x-4)$$
and for $x=2$ this is 8, but for $x=3$ this is 0. Furthermore, this expression is non-negative over $[2,3]$. Therefore, since $2<a_2<3$, the sequence monotonically converges to 3 from below after $a_1$, so $m=2$.
A: Alt. hint: quite obviously $\;a_n \ne 0\,$, then inverting both sides gives:
$$
\frac{1}{a_{n+1}}=1 - 4 \frac{1}{a_n} + 6 \frac{1}{a_n^2} \quad\iff\quad b_{n+1} = 6 b_n^2 - 4b_n + 1 \quad \style{font-family:inherit}{\text{where}} \;\; b_{n} = \frac{1}{a_n}
$$
Multiplying by $\,6\,$ then gives:
$$
6 b_{n+1} = 36 b_n^2 - 24b_n + 6 = (6 b_n - 2)^2+2 \quad\iff\quad c_{n+1} = c_n^2 \quad \style{font-family:inherit}{\text{where}} \;\; c_{n} = 6 b_n - 2
$$
Since $|c_1| = | 6 / a_1 - 2| = 4/5 \lt 1$ it follows that $\,c_n \to 0\,$ from above, so $\,a_n \to 3\,$ from below, then the same argument already made in other answers applies to show that $\,m = 2\,$.
