Do factorials really grow faster than exponential functions? [closed]

Having trouble understanding this. Is there anyway to prove it?

Solution 1:

If you're not quite in the market for a full proof:

$$a^n=a\times a\times a\times a...\times a$$ $$n!=1\times 2\times 3\times 4...\times n$$

Now what happens as $n$ gets much bigger than $a$? In this case, when $n$ is huge, $a$ will have been near some number pretty early in the factorial sequence. The exponential sequence is still being multiplied by that (relatively tiny) number at each step, while $n!$ is being multiplied by $n$. So even if $n!$ starts out small, it'll eventually start being multiplied by gigantic numbers at each step, and quickly outgrow the exponential. If $a=10$ and $n=100$, then $a^n$ has around $100$ digits, while $n!$ has over $150$ digits. Note that near $n=100$, $n!$ is having roughly 2 digits added per step (and that rate will only increase), while $a^n$ is still only ever going to get one more with every step. No contest.

Solution 2:

Let me give a Hint: Let $f(n) = \dfrac{n! }{ a^n}$, for $ a > 1$. What is $\dfrac{f(n+1)}{f(n)}$??

Solution 3:

An intuitive way to see this is to consider that you're trying to show $$a^n < n!$$ for sufficiently large $n$. Take the log of both sides, you get $$n\log(a) = \log(a^n) < \log(n!) = \sum_{i = 1}^n\log(i).$$ Now as you increase $n$ you only add $\log(a)$ to the left side, but the $\log(n + 1)$ that you add to the right can be arbitrarily large as $n$ becomes large. This can be made rigorous, but I think that it's intuitively clear that eventually it gets large enough to make up the difference and be greater than $n\log(a)$.

Solution 4:

Why does the function $exp(x)$ converge?

Since

$$\exp(x)=\sum_{i=0}^{\infty} \frac{x^{n}}{n!}$$ for large $n$, $x^n$ grows slowly compared with $n!$.