Calculating the Limit of (n)^(1/n)

[lockedup]

Homework Statement

Calculate lim (n!)^1/n

Homework Equations

The Attempt at a Solution

Call the limit L.

lim ln(n!)^1/n = ln L

ln(n!)^1/n = \frac{1}{n}ln(n!) = \frac{1}{n}(ln(1) + ln(2) + ... + ln(n)) = \infty/n. So the limit is \infty/0? Which makes L what?

This is for Analysis and we haven't got to series yet and I took calc II eight years ago.

 

[algebrat]

seems you need to head towards l'Hopital's rule. and you need to get rid of that factorial. Stirling's formula helped me get closer to the idea, but you don't need to know that. you were on the right track, just can't have the number of terms changing is the only thing that's obstructing your current strategy

go for ln(n!)<n*ln(n)

by the way, you didn't have ∞/0, you had ∞/∞, but the number of terms was changing, so you couldn't use l'Hospital's rule

 

[D H]

Does your book or your notes say anything about Stirling's approximation?

 

[lockedup]

I've never heard of Stirling's Approximation. This problem is tacked on the end of the section on lim sup's and lim inf's.

 

[D H]

You should be able to show that this an unbounded expression. It does however have an asymptotic behavior that can be expressed quite compactly.

 

[Bohrok]

What if you use the fact that n! > aⁿ for all real a, then (n!)^1/n > a?

 

[algebrat]

[STRIKE]i see none of you read my first response above... it is gospel, i am prophet, the year is 2012, i told you so[/STRIKE]

 

[algebrat]

What if you use the fact that n! > aⁿ for all real a, then (n!)^1/n > a?

[STRIKE]your a depends on n, it will not work[/STRIKE]

EDIT: er, no it doesn't but something doesn't look right, the onus is on me now?

 

[algebrat]

[STRIKE]your a depends on n, it will not work[/STRIKE]

EDIT: er, no it doesn't but something doesn't look right, the onus is on me now?

I tap out, tired

 

[lockedup]

I think I got it. I looked through my old calc book. I found this little fact: \intln x dx = x ln x - x + 1 < ln (n!)

 

[D H]

go for ln(n!)<n*ln(n)

That won't work because all this will say is that (n!)^1/n might be bounded. A lower limit is needed. You gave an upper limit.

What if you use the fact that n! > aⁿ for all real a, then (n!)^1/n > a?

Plugging in n=1, a=2 yields 1>2, which is obviously false.

I think I got it. I looked through my old calc book. I found this little fact: \intln x dx = x ln x - x + 1 < ln (n!)

You didn't show the limits, but yep, you're on the right track. This is a good lower bound.

 

[algebrat]

What if you use the fact that n! > aⁿ for all real a, then (n!)^1/n > a?

Plugging in n=1, a=2 yields 1>2, which is obviously false.

Just to prove that I can be helpful: I think Bohrok meant that for any a, choose n large enough, which is fine since n->∞.

 

[Bohrok]

Just to prove that I can be helpful: I think Bohrok meant that for any a, choose n large enough, which is fine since n->∞.

Thank you

It was late for me when I wrote that; what I meant to write was:
What if you use the fact that, for sufficiently large n, n! > aⁿ for all real a, then (n!)^1/n > a?

That still may not be the best way of wording it, but you get the idea.


Here's another thread about the same limit.
The OP has an interesting approach (which he didn't finish), but I can't figure out where his teacher was going after that point.

 

[algebrat]

Thank you

It was late for me when I wrote that; what I meant to write was:
What if you use the fact that, for sufficiently large n, n! > aⁿ for all real a, then (n!)^1/n > a?

That still may not be the best way of wording it, but you get the idea.Here's another thread about the same limit.
The OP has an interesting approach (which he didn't finish), but I can't figure out where his teacher was going after that point.

Glad I could help. But let me offer a resolution to your wording. When I took a class on proofs, I remember the instructor, Steffen Rohde, mentioned that saying that this holds for all x could run you into trouble. It is safer to say, for all x, this holds.

Like the difference between f(x) \forall x, and \forall x, f(x). The second one comes off less ambiguously.

When proofs get really confusing, I sort of imagine I'm in a subset after I fix some variable, whether it is by "there is" or "for any".

In the end, I think that's sort of on D H for not interpreting that "naturally". I like to use the word natural in math for when some there is an obvious way to take things.