# Lim n^n/n! as n approaches infinity

1. Oct 9, 2016

### Battlemage!

1. The problem statement, all variables and given/known data
$$\lim_{n\rightarrow ∞}\frac{n^n}{n!}$$

2. Relevant equations
n! = (1)⋅(2)⋅(3)⋅...⋅(n-1)⋅n

3. The attempt at a solution
$$\lim_{n\rightarrow ∞}\frac{n^n}{n!}$$
$$\lim_{n\rightarrow ∞}\frac{n^n}{(1)⋅(2)⋅...⋅(n-1)⋅n}$$

I then factor n out of the denominator n times, or rather, nn, leaving:

$$\lim_{n\rightarrow ∞}\frac{n^n}{n^n⋅(\frac{1}{n})⋅(\frac{2}{n})⋅...⋅(1-\frac{1}{n})⋅(1)}$$
$$\lim_{n\rightarrow ∞}\frac{1}{(\frac{1}{n})⋅(\frac{2}{n})⋅...⋅(1-\frac{1}{n})⋅(1)}$$
which is:
$$\frac{1}{0⋅0⋅...⋅(1-0)} = \frac{1}{0}=∞$$

This is the result Wolfram gives, but I can't get the step by step since I don't have Pro.
http://www.wolframalpha.com/widgets/view.jsp?id=7c220a2091c26a7f5e9f1cfb099511e3

Is this acceptable, or if not is there an alternative way to do this?

2. Oct 9, 2016

### Ray Vickson

When you ask if "this is acceptable", what do you mean by "this"? If you are asking whether it is acceptable to use Wolfram Alpha to replace analysis and thinking, then I would say no. If you are asking whether it is acceptable to say
$$\lim_{n \to \infty} \frac{1}{(\frac{1}{n})⋅(\frac{2}{n})⋅...⋅(1-\frac{1}{n})⋅(1)} = \frac{1}{0⋅0⋅...⋅(1-0)}$$
the answer is no again. Dividing by zero is never, ever, acceptable under any circumstances; it is one of the fundamental taboos of mathematics.

Last edited: Oct 9, 2016
3. Oct 9, 2016

### Staff: Mentor

The "..." are a bit problematic. You are basically taking two limits at the same time here, and in general that is not well-defined, although it works here.

I would just keep the first 1/n and find an upper limit for the remaining product, which then gives a lower limit on n^n/n! for every n.

4. Oct 10, 2016

### Battlemage!

What I meant is "does not exist." The limit blows up to infinity so it cannot be a limit.

That mistake aside, what about the steps I used to get to that point? For example, is factoring out nn from the denominator acceptable, and if so did I do it right?

*as for using Wolfram, I use it only as a means of checking my work. As I describe below, such tools and forums like this are all I have at this moment.

I will look into this.

Thanks both of you. I'm trying to learn and relearn this stuff on my own so I can get back to school with a sharper mind (had to quit a while due to personal circumstances), so it helps to have some advise on these problems on the forums. All the insight is very valuable to me.

5. Oct 10, 2016

### Staff: Mentor

That is fine.

6. Oct 10, 2016

### Ray Vickson

Your argument is basically OK, but can be fixed up a bit. However, you are doing it the hard way. Start with
$$\frac{n^n}{n!} = \frac{n}{n} \frac{n}{n-1} \cdots \frac{n}{2} \frac{n}{1}$$
We have $n/n=1, n/ (n-1) > 1, \ldots, n/2 > 1$, so $n^n/n! > 1 \cdot 1 \cdot 1 \cdots \cdot 1 \cdot n = n$; that is, $n^n/n! > n$, so $n^n/n \to \infty$ as $n \to \infty$.

Last edited by a moderator: Oct 10, 2016
7. Oct 10, 2016

### Battlemage!

Would this be some form of the squeeze theorem?

8. Oct 10, 2016

### Staff: Mentor

It is just one-sided here, but it has some similarity.

9. Oct 10, 2016

### Ray Vickson

I would hardly dignify it with the term "theorem"; it is just the obvious statement that if $t(n) > n$ then $t(n) \to \infty$ as $n \to \infty$.