Potential and factorial limit that tends to infinity

Question

I've a question about this limit. As you can see it has a potential argument and a factorial one as well.

$$\frac{n^p}{n!} $$ ($p$ belongs to $N_1$)

When the limit tends to infinity, the fraction stays like this $\frac{\inf}{\inf}$

But because the factorial is bigger than the $n^p$, the limit tends to zero. My question is: how can I prove mathematically that the factorial term is bigger than the potential one?

Thank you so much for your help. If something is not very clear, please let me know.

Answer 1

Let $a_n:=\frac{n^p}{n!}$, then $\frac{a_{n+1}}{a_n}=(1+\frac{1}{n})^p \frac{1}{n+1} \to 0$ as $ n \to \infty$. Hence the series $\sum_{n \ge 1}a_n$ is convergent. Therefore $a_n \to 0$.

Answer 2

Expanding on JMoravitz's comment, take the natural log to get $\ln(n^p/n!)$. Simplifying, we get $\ln(n^p) - \ln(n!) = p \ln n - \ln n!$, then we use the approximation $\ln n! \approx n \ln n - n + o(\ln n)$ to get $(p-n) \ln n - n$.

Now find the roots of $(p-n) \ln n - n$. Will this approximation have a root?