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
<br /> \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 &lt; 1(D'Alembert test),

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

Then

<br /> \lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.<br />

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
<br /> \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 &lt; 1(D'Alembert test),

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

Then

<br /> \lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.<br />

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