# Limit of $\frac{n!}{2^{(n^{2})}}$

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 $$a\in\Bbb{N}$$?

• $2^{n^2}$ grows faster than $n!$ – Jeremy Oct 3 at 17:58
• OP, l’Hopital on each factor as you have written it in your post (as it is indeterminate) shows that they identically tend to 0. For terms with a finite numerator it is automatically 0. Perhaps try to flesh this out. – Certainly not a dog Oct 3 at 18:13
• It seems incorrect. I edited my first post. I checked on Wolfram. – Jeremy Oct 3 at 18:15

It is easy to see that $$n/2^n \leq 1/2$$ for each integer $$n \geq 1$$.

So we can limit $$a_n = \frac{n!}{2^{n^2}}$$ (from above) with a geometric progression, I think.

How? Prove the above inequality by induction.

Then using that inequality we have:

$$\frac{n!}{2^{n^2}} \leq \frac{n^n}{2^{n^2}} = (\frac{n}{2^n})^n \leq (1/2)^n$$

Then we can find that the limit is zero using the squeeze theorem.

• A very concise and clear answer! – NotAMathematician Oct 4 at 20:08

$$n! \leq n^n=2^{n\log_2 n}$$ This should give you your squeeze theorem argument.

We can prove directly by induction proving that $$n!\le2^{n^2-2n+1}$$, indeed

• base case: $$n=1 \implies 1\le 1$$

• induction step: we need to prove

$$k!\le2^{k^2-2k+1}\implies (k+1)!\le2^{k^2}$$

which is true indeed

$$(k+1)!=(k+1)k!\le(k+1)2^{k^2-2k+1}\le2^{k^2-2k+2}\le 2^{k^2}$$

then

$$n!\le2^{n^2-2n+1}< 2^{n^2}$$

and then

$$\frac{n!}{2^{n^2}}\le \frac1{2^{2n-1}}< 1$$

By ratio test

$$\frac{\frac{(n+1)!}{2^{(n+1)^2}}}{\frac{n!}{2^{n^2}}}=\frac{n+1}{2^{2n+1}}\to 0$$

and the latter can be easily proved by induction or ratio test again.

Edit: Note that ratio test is essentially equivalent to the squeeze theorem ( see the proof).

Seemingly a bit late this answer but I think it is worth mentioning it.

A possible way is as follows:

• Consider first $$a_n = \sqrt[n]{\frac{n!}{2^{n^2}}}=\frac{\sqrt[n]{n!}}{2^n}\stackrel{GM-AM}{\leq} \frac{\frac{n(n+1)}{2n}}{2^n} = \frac{n+1}{2^{n+1}}\stackrel{2^{n+1}=(1+1)^{n+1}}{\leq}\frac{n+1}{\binom{n+1}{2}}=\frac{2}{n}\stackrel{n\to \infty}{\longrightarrow}0$$

Since $$\lim_{n\to \infty}a_n = 0 \Rightarrow \lim_{n\to \infty}a_n^n = 0$$.

Your general case for $$a\in\mathbb{N}, a \geq 2$$ follows immediately from the case $$a=2$$, since $$\frac{n!}{a^{n^a}}\leq \frac{n!}{2^{n^a}}\leq \frac{n!}{2^{n^2}}$$.