Squeeze Theorem with limits n/n as n approaches 0

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Homework Help Overview

The question involves using the squeeze theorem to demonstrate that the limit of n!/n^n approaches 0 as n approaches infinity. Participants are exploring the application of the theorem and the identification of upper and lower bounds.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss potential upper limits and lower limits for the expression n!/n^n, with some suggesting specific values and others questioning how to generalize the findings across different values of n.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants are attempting to clarify the reasoning behind the bounds, while others are exploring the implications of specific cases, such as n=6. There is no explicit consensus, but multiple lines of reasoning are being examined.

Contextual Notes

Participants are working within the constraints of the problem statement and are unsure about certain definitions and setups related to the squeeze theorem. There is a focus on understanding how to apply the theorem without reaching a final conclusion.

ReginaldN
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Homework Statement


The question asks to use the squeeze theorem to show that the limit of n!/n^n equals 0 as n approaches ∞.




Homework Equations


I need to use the squeeze theorem to solve this problem, and I'm not sure what the upper limit is.


The Attempt at a Solution


I found the lower limit to be 1/n^n, and I don't remember how I got here. I need some help.
 
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Isn't it just n^1? Which is just n?
 
You can write n!/n^n as (1/n)*(2/n)*(3/n)*...*((n-1)/n)*(n/n), right? All of those factors are less than or equal to 1. How many of those factors are less than or equal to 1/2?
 
hmm, isn't 2/2^2 equal to 1/2? So only one of those is less than or equal to 1/2?
 
ReginaldN said:
hmm, isn't 2/2^2 equal to 1/2? So only one of those is less than or equal to 1/2?

No, take n=6. That's 6!/6^6=(1/6)*(2/6)*(3/6)*(4/6)*(5/6)*(6/6). Isn't it? I think three of those are less than or equal to 1/2. The first three. Can you generalize? n odd is a little different, but don't worry about that right now.
 
okay so, for n=6. everything less than 3/6 is less than or equeal to one half? I'm not really understanding what you mean by generalize.
 
ReginaldN said:
okay so, for n=6. everything less than 3/6 is less than or equeal to one half? I'm not really understanding what you mean by generalize.

Generalize means how many for n=10, n=20 etc. Now how many for a general even value of n?
 
So for about half of the values of n, half of them will be great than or equal to 1/2, right?
 
Can you show n!/(nn) ≤ (1/n) ?
 
  • #10
deluks917 said:
Can you show n!/(nn) ≤ (1/n) ?

deluks917 is right and has a much simpler approach. I was trying to get you to show n!/n^n<=(1/2)^(n/2). That works but you don't need that much unless you are trying to show the series converges. n!/n^n<1/n will do just fine. Can you say why that must be true?
 
  • #11
that must be true, because no matter what integer n is, it will always end up as 1/n. For example, if n = 2 , then the equation would be 2/22, which is 21 / 22, and that would equal 1/21.
 

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