Prove that $y_n$ = $\frac{6^n}{n!}$ is contractive Prove that $y_n$ = $\frac{6^n}{n!}$ is contractive.
My attempt at this question is to compare $\frac{|y_{n+2}-y_{n+1}|}{|y_{n+1}-y_{n}|}$. Doing this got me to $\frac{|24-6n|}{|-n^2+3n+10|}$, but I don't know how to get a constant to prove that it is contractive.
 A: In my opinion, the sequence is not contractive. A short answer could be that the polynomial at the denominator has a root for $n=5$, so the value of the fraction is not defined (or it is $\infty$, anyway it is not $\lt 1$). A different way could be by observing that the sequence has the same value for 5 and 6, so we cannot find a real number c,  $0\lt c \lt 1$ such as $ \vert y_7 - y_6 \vert \le 0c=0$, since the number on the left side is not 0.
A: $$\left| \frac{y_{n+2}-y_{n+1}}{y_{n+1}-y_n}\right|\le 1$$
simplified gives
$$\frac{n-4}{(n-5) (n+2)}\leq \frac{1}{6}$$
which is verified, provided that $n\ge 7$.
Edit
Thus it is not contractive since the inequality is not true for any $n\ge 1$.
It can easily made contractive shifting $n$ as follows
$$y_n=\frac{6^{n+7}}{(n+7)!}$$
A: $\frac{|24-6n|}{|-n^2+3n+10|}$
= $\frac{6|4-n|}{|-n^2+3n+10|}$
$<=$ $\frac{|4-n|}{|-n^2+3n+10|}$
$=$ $\frac{|n-4|}{|-n^2+3n+10|}$
$<=$ $\frac{|n|}{|-n^2+3n+10|}$
$<=$ $\frac{|n|}{|3n|}$
=  $\frac{1}{3}$
0 < 1/3 < 1 , so we have a contractive sequence
