What is the Limit of (1/n!)^(1/3n) as n Approaches Infinity?

[Dschumanji]

Homework Statement

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.

Homework Equations

The Squeeze Theorem

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?

 

[hunt_mat]

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

 

[Dschumanji]

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

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?

 

[SammyS]

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)}\ ?

 

[Dschumanji]

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)}\ ?

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.

 

[SammyS]

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.

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

 

[Dschumanji]

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

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

 

[SammyS]

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.

 

[Dschumanji]

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.

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

 

[SammyS]

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

 

[Dickfore]

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

 

[Dschumanji]

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

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