Steps to find $\lim_\limits{n\to +\infty} \frac{n^n+3n!}{n^n+(3n)!} = 0$

Question

I don't understand why $\lim_\limits{n\to +\infty} \frac{n^n+3n!}{n^n+(3n)!} = 0$

It's taken from Basisboek wiskunde (8.22d) from Jan van de Craats.

The solution given is:

$0$ (because $(3n)! = n! \times (n+1)(n+2)...3n>n! \times n^n$, etc.)

Actually, I don't understand this answer at all, I'd need some more steps to get it, could someone help me to make it clearer ?

Answer 1

We have

$(3n)! = n! \times (n+1)(n+2)...(3n) > n! \times \underbrace{(n)(n)...(n)}_{\text{2n times}} = n! \times n^{2n} \geq n! \times n^n$.

The key step above is replacing $n+1$ , $n+2$ , $n+3$ etc with $n$ each time.

Thus we have:

$$\frac{n^n+3n!}{n^n+(3n)!} < \frac{n^n+3n!}{n^n + (n! \times n^n)}$$

Can you finish it up from here?