Find the following limit:$$\lim_{n\to\infty} \frac{n!}{2^{n^2}}$$
I get the feeling that this limit equals to zero. Intuitively, the function $f(n)=2^{n^2}$ grows much more faster than the factorial, however, I wish to prove this limit using only the squeeze theorem or some algebra. I noticed that:
$$0\leq \frac{n!}{2^{n^2}}=\frac{n}{2^n} \cdot\frac{n-1}{2^n}\cdots \frac{2}{2^n}\cdot\frac{1}{2^n}$$
I tried to think about cases about wether $n$ is even or odd, hoping that would lead me to a way to simplify the latter expression, but it didn't work. Also, is there a way to generalize the problem? That is, does the limit
$$\lim_{n\to\infty} \frac{n!}{a^{n^a}}$$ is always equal to zero, for $a\in\Bbb{N}$?
Thanks in advance!
Best Answer
$$n! \leq n^n=2^{n\log_2 n} $$ This should give you your squeeze theorem argument.