Why Does n! Grow Faster Than a^(2n+1) as n Approaches Infinity?

[physicsRookie]

Show that
$$\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \infty \qquad a \in \mathbb{N}$$


Here is my try:

Take $$a_n = a^{2n+1}/n!$$ . Since
$$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a^{2(n+1)+1}/\,(n+1)!}{a^{2n+1}/\,n!} = \lim_{n \to \infty} \frac{a^{2}}{n+1} = 0 < 1$$(D'Alembert test),

the series $$\sum_{n=0}^\infty a_n$$ is convergent, i.e.$$\lim_{n \to \infty} a_n = 0.$$

Then

$$\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.$$

Correct? Is there another way to prove it?
Thanks.

 

[topsquark]

Show that
$$\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \infty \qquad a \in \mathbb{N}$$Here is my try:

Take $$a_n = a^{2n+1}/n!$$ . Since
$$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a^{2(n+1)+1}/\,(n+1)!}{a^{2n+1}/\,n!} = \lim_{n \to \infty} \frac{a^{2}}{n+1} = 0 < 1$$(D'Alembert test),

the series $$\sum_{n=0}^\infty a_n$$ is convergent, i.e.$$\lim_{n \to \infty} a_n = 0.$$

Then

$$\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.$$

Correct? Is there another way to prove it?
Thanks.

That works. You can also approach it from the standpoint that n! is eventually larger than any power of fixed a for n large enough. (Well, as long as we aren't talking about $$a^{n!}$$ or something crazy like that!)

-Dan