Proving the Limit of an Expression: \frac{n^n}{n!}

  • Thread starter Thread starter ercagpince
  • Start date Start date
  • Tags Tags
    Expression Limit
Click For Summary

Homework Help Overview

The discussion revolves around finding the limit of the expression \(\lim_{n\rightarrow\infty}\frac{n^n}{n!}\). Participants are exploring methods to prove the limit mathematically, with some suggesting the use of Stirling's approximation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches, including the application of Stirling's approximation and the potential use of logarithms. There are questions about the necessity of taking the logarithm of the numerator and requests for explicit demonstrations of the methods suggested.

Discussion Status

The discussion is active, with participants offering different perspectives on how to approach the problem. Some guidance has been provided regarding the use of Stirling's approximation, but there is no explicit consensus on the best method to proceed.

Contextual Notes

There are indications of confusion regarding the application of Stirling's approximation and the role of logarithms in the limit evaluation. Participants are also navigating the expectations of providing detailed steps in their reasoning.

ercagpince
Messages
29
Reaction score
0

Homework Statement


what is the limit of this expression?

[tex]lim_{n\rightarrow\infty}\frac{n^n}{n!}[/tex]

Homework Equations


The Attempt at a Solution


I tried to make it look like [tex]\frac{x^n}{n!}[/tex] and also tried to apply the sandwich theorem, but got nothing logical.
Probably the limit is [tex]\infty[/tex], still I want to prove it mathematically.
 
Physics news on Phys.org
You could take the log and use Stirling's approximation on the factorial.
 
Could you show it explicitly?
 
ercagpince said:
Could you show it explicitly?

If you mean show the details, Isn't that your job? Did you look up Stirling's approximation? Or do you mean do it without Stirling's formula?
 
if you take the stirling's approximation there is no need to take log of the term on numerator.
that why i asked that.
 
ercagpince said:
if you take the stirling's approximation there is no need to take log of the term on numerator.
that why i asked that.

Right, if you use the form n!~(n/e)^n. I was thinking of ln(n!)~n*ln(n)-n
 
thanks for the post
 

Similar threads

Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K