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.

$$\left| \frac{y_{n+2}-y_{n+1}}{y_{n+1}-y_n}\right|\le 1$$
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
$$y_n=\frac{6^{n+7}}{(n+7)!}$$
