# Homework Help: A Hard Factorial Limit

1. Jul 3, 2011

### Dschumanji

1. The problem statement, all variables and given/known data
I need to evaluate the following limit:

$lim_{n\rightarrow\infty}\left(\frac{1}{n!}\right)^{\frac{1}{3n}}$

Now, I think I have solved it, but my methods are not rigorous. I am looking to see if my assumptions are sound and if there is an easier way to solve this limit.

2. Relevant equations
The Squeeze Theorem

3. The attempt at a solution
My main approach to solving this limit is to use the squeeze theorem. I think it is safe to assume that the the expression must be greater than or equal to zero; this provides my lower bound. I want to try and find an upper bound with the following form:

$\left(\frac{1}{n}\right)^{\frac{1}{b}}$

Where b is some fixed integer. By letting this be the upper bound, I end up with the following inequalities:

$0 \leq \left(\frac{1}{n!}\right)^{\frac{1}{3n}} \leq \left(\frac{1}{n}\right)^{\frac{1}{b}}$

The problem is to find an integer b that makes the above inequalities true for some value of n greater than 1. This required a bit of guessing and checking since I have no idea how to find exact values to the following inequality (it is the one immediately above involving the expressions with exponenets except that it is rewritten):

$n!^{\frac{b}{3}} - n^{n} \geq 0$

When I let b = 4, the inequalities become true at around n = 50. This should let me apply the squeeze theorem to conclude that the limit must equal zero, right?

2. Jul 3, 2011

### hunt_mat

How about looking at $n!\leqslant n^{n}$? What does that tell you?

3. Jul 3, 2011

### Dschumanji

I did use a similar inequality to determine that b cannot be equal to 1, 2, and 3, but it doesn't help in determining if there exists an n greater than 1 such that every n after makes $n!^{\frac{4}{3}}-n^{n} \geq 0$ a true statement.

Now, if I used the inequality $n! \leq n^{n}$ to help with finding an upper bound you end with the following:

$\left(\frac{1}{n!}\right)^{\frac{1}{3n}} \geq \left(\frac{1}{n^{n}}\right)^{\frac{1}{3n}}$

Before simplifying, it is clear that $n! \leq n^{n}$ is not a very helpful fact for finding an upper bound since the expression which is supposed to be an upper bound becomes a lower bound. In fact, after simplifying you find out that the right hand side of the inequality turns into the form of upper bound I was originally looking for when b is equal to 3. It shows that b cannot be equal to 3.

This is what I see, am I missing something?

4. Jul 3, 2011

### SammyS

Staff Emeritus
What about comparing $\displaystyle \left(\frac{1}{n!}\right)^{1/(3n)} \text{ and } \left(\frac{1}{n}\right)^{1/(3n)}\ ?$

or $\displaystyle \left(\frac{1}{n!}\right)^{1/(3n)} \text{ and } \left(\frac{1}{3n}\right)^{1/(3n)}\ ?$

5. Jul 3, 2011

### Dschumanji

If you compare the first two, the second expression approaches 1 as n increases without bound. If you compare the next two, the second expression approaches 1 as n increases without bound. If you examine the expression in the original limit, you can see that it does not approach one, it dips down below 1 and slowly approaches what looks like zero. Both of the suggested upper bounds are not useful for the squeeze theorem.

Last edited: Jul 3, 2011
6. Jul 3, 2011

### SammyS

Staff Emeritus
Yup! Silly mistake on my part.

b = 4 does work, with $(n!)^{\frac{b}{3}} - n^{n} \geq 0$, for n > 42 .

However, b = 6, gives $\displaystyle (n!)^{\frac{b}{3}} - n^{n} \geq 0$, for n > 2 , and it's pretty easy to show this.

Also, $\displaystyle \lim_{n\to\infty}\left(\frac{1}{n}\right)^{1/6}=0$

7. Jul 3, 2011

### Dschumanji

How do you go about solving for n when a specific b is input into the inequality?

8. Jul 3, 2011

### SammyS

Staff Emeritus
Numerically. But $\displaystyle (n!)^{2} - n^{n} = 0$ is true for n = 1, 2.

For b = 4, I used WolframAlpha. Then actually plugged in n=41, which gave a negative answer & n = 42 which gave a positive.

9. Jul 4, 2011

### Dschumanji

Ah, thank you so much SammyS! And thank you everyone else! Letting b equal 6 is definitely much better than letting b equal 4.

10. Jul 4, 2011

### SammyS

Staff Emeritus
You're welcome! & I apologize again for my mistaken first suggestion.

11. Jul 4, 2011

### Dickfore

Rewrite it as:
$$\left(\frac{1}{n!}\right)^{\frac{1}{3 n}} = \exp{\left[-\frac{\ln{(n!)}}{3 n}\right]}$$
and use Stirling's asymptotic formula:
$$\ln{(n!)} \sim \left(n + \frac{1}{2}\right) \, \ln{n} - n, \ n \rightarrow \infty$$

12. Jul 4, 2011

### Dschumanji

This method is much simpler! I love it! Thank you so much, Dickfore!

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