Lim n^n/n as n approaches infinity

[Battlemage!]

Homework Statement

$$\lim_{n\rightarrow ∞}\frac{n^n}{n!}$$

Homework Equations

n! = (1)⋅(2)⋅(3)⋅...⋅(n-1)⋅n

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=7c220a2091c26a7f5e9f1cfb099511e3Is this acceptable, or if not is there an alternative way to do this?

 

[Ray Vickson]

Homework Statement

$$\lim_{n\rightarrow ∞}\frac{n^n}{n!}$$

Homework Equations

n! = (1)⋅(2)⋅(3)⋅...⋅(n-1)⋅n

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=7c220a2091c26a7f5e9f1cfb099511e3Is this acceptable, or if not is there an alternative way to do this?

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.

 

[mfb]

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.

 

[Battlemage!]

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.

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.

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.

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.

 

[mfb]

For example, is factoring out nⁿ from the denominator acceptable, and if so did I do it right?

That is fine.

 

[Ray Vickson]

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.

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 $n \to \infty$.

Edit (6/1/22): The above has a typo. I'm sure Ray meant that is, $n^n/n! > n$, so ##n^n/n! \to \infty$as$n \to \infty##.

 

[Battlemage!]

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 $n \to \infty$.
Edit (6/1/22): The above has a typo. I'm sure Ray meant that is, $n^n/n! > n$, so $n^n/n! \to \infty$ as $n \to \infty$.

Would this be some form of the squeeze theorem?

 

[mfb]

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

 

[Ray Vickson]

Would this be some form of the squeeze theorem?

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 $n \to \infty$.