$\displaystyle \frac {n^3}{n!} \to 0 $
Hi there, I would expand this quoient out and see what happens
$\displaystyle \frac {n^3}{n!} = \frac{n \times n \times n}{n\times (n-1) \times (n-2)\times (n-3)\times \dots}= $ $\displaystyle \frac{n \times n }{ (n-1) \times (n-2)\times (n-3)\times \dots} = \dots$
Hi!
I would check that $\displaystyle n^3 = n(n-1)(n-2) + 3n(n-1) +n$.
This gives
$\displaystyle \frac{n^3}{n!}= \frac{ n(n-1)(n-2)}{n!} +\frac{ 3n(n-1)}{n!} + \frac{n}{n!} = \frac{1}{(n-3)!} +\frac{3}{(n-2)!} +\frac{1}{(n-1)!}$,
each summand clearly goes to zero.