How can you proof that $\lim_{n \to \infty} n!=\infty$?

Question

When solving a limit of a succession last class involving $n!$, the professor said we could proof that $\lim_{n \to \infty} n!$ is $\infty$, but he left it as some sort of homework the actual process of the proof.

So far I tried an induction approach, since the first terms of the succession (I'm using the recursive defintion of $a_n=na_{n-1}$) are greater than the previous, I stated that $$n!>(n-1)!$$ and for $n+1$ then $$(n+1)!=(n+1)n!$$ And since $n+1>0$ $$(n+1)n!>n!$$ The thing is that I'm not sure about how should I finish the proof, is there some sort of $\delta-\epsilon$ process involved?

Thanks for reading

Answer 1

Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $\delta$-$\epsilon$, but $N$-$\epsilon$ (we're working with sequences here, as opposed to functions $\mathbb{R} \to \mathbb{R}$). Furthermore, because the limit is not finite, you shouldn't be using $\epsilon$.

To clarify, we want to show that $$\lim_{n \to \infty}n! = +\infty\text{.}$$ To show this, the definition is as follows:

$\lim_{n \to \infty}n! = +\infty$ means:

For every $\alpha \in \mathbb{R}$ there is a positive integer $N$ such that for all $n > N$, $n! > \alpha$.

See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.

To show this, we may approach as follows.

Let $\alpha \in \mathbb{R}$ be arbitrary.

Suppose $\alpha < 1$ and let $N = 1$. Then for each $n \geq 1$, $$n! = n(n-1)\cdots1\geq \underbrace{1 \cdot 1 \cdots 1}_{n \text{ times}} = 1^n=1> \alpha\text{.}$$ Now suppose $\alpha \geq 1$. If $\alpha \geq 1$, take $N = \lceil\alpha\rceil+1$, so that for each $n \geq N$, $$n! = n(n-1)\cdots1\geq n\geq N=\lceil\alpha\rceil+1\geq\alpha+1>\alpha$$ as desired.

Answer 2

It suffices simply to note that

$$n!=n(n-1)\ldots 2\cdot 1\ge n \to \infty$$

Answer 3

The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $\ln n!=\sum_{k=1}^n\ln k\to\infty$ as $n\to\infty$; this follows from $\ln k\ge \ln 2>0$ for $k\ge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!\ge 2^{n-1}$ for $n\ge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1\ge 2$.)

Answer 4

Hint:

$$n! \geq n$$ for all $n\in\mathbb N$.