Result of $\lim_{n\to\infty} \frac{{x}^{100n}}{n!}$

Question

The task is to $\lim_{n\to\infty} \frac{{x}^{100n}}{n!}$. n is an integer. I've tried to use Stolz theorem, but that doesn't seem to give any result.

Thank you for your help.

Answer 1

By ratio test

$$\frac{{x}^{100(n+1)}}{(n+1)!}\frac{n!}{{x}^{100n}}=\frac{x^{100}}{n+1}\to 0$$

then

$$\lim_{n\to\infty} \frac{{x}^{100n}}{n!}=0$$

Answer 2

Letting $y=x^{100}$ you want $\lim_{n\to\infty} \frac{y^n}{n!}.$

Let $z_n=\frac{|y|^n}{n!}.$ Then $z_{n+1}=\frac{|y|}{n+1}z_n$.

Now, when $n+1\geq 2|y|$ you have that:

$$0\leq z_{n+1}\leq \dfrac{1}{2}z_n$$

By induction, you get $z_{n+k}\leq \frac{1}{2^k}z_{n}$.

So $z_{n}\to 0.$

Answer 3

Hint:

$$\exp(x^{100}) = \sum_{n=0}^\infty \frac{(x^{100})^n}{n!} = \sum_{n=0}^\infty \frac{x^{100n}}{n!}$$

so $$\lim_{n\to\infty} \frac{x^{100n}}{n!} = 0$$

Answer 4

Note that $x^{100n} = 2^{100n \ln x} = 2^{100a n}$ for some constant $a$ (which may be 0 or even negative).

$n! \approx \frac{n^n\sqrt{2\pi n}}{e^n}$ (Stirlings) which is $2^{\Omega(n \log n)}$.

Note that $\frac{2^{100an}}{2^{\Omega(n \log n)}}$ goes to 0 for any constant $a$ (including $a \le 0$). This answers your question.