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Limit of a sequence n!/n^n

  • Thread starter Tomp
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  • #1
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Homework Statement

Using the sandwich rule (which i understand) find the limit of n!/nn

The attempt at a solution

To my knowledge n! is the fastest growing function you can have, so I immediately thought the function did not have a limit, however, the answer states the limit to be 1

I know that the bound on n! would be 1 =< n! < inf (?) but then dividing through by nn doesn't really help.

I am stuck :/

What am I doing wrong?
 

Answers and Replies

  • #2
24
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FALSE ALARM

After a bit of searching on the internet, the answer in the book is wrong. phew!

Answer is 0 as i initially thought
 
  • #3
DryRun
Gold Member
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What limits did you compare it to, for the Sandwich theorem?
 
  • #4
880
40
Homework Statement

Using the sandwich rule (which i understand) find the limit of n!/nn
Just asking what the limit of f(x) is, gives no meaning. You need to specify to what value n approaches?
 
  • #5
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Just asking what the limit of f(x) is, gives no meaning. You need to specify to what value n approaches?
apologies as n approaches infinity
 
  • #6
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What limits did you compare it to, for the Sandwich theorem?
i said the lower limit for n! was 1 and the upper limit was infinity....
 
  • #7
880
40
apologies as n approaches infinity
Aye, 0 is correct, then. :smile:

If, supposedly, n->0, then the limit would be 1 :wink:

To my knowledge n! is the fastest growing function you can have
From this reasoning(incorrect), I don't see how you got 0 as your answer though....
 
  • #8
DryRun
Gold Member
838
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i said the lower limit for n! was 1 and the upper limit was infinity....
And how did you end up with those limits? Just curious. :smile:
 
  • #9
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Homework Statement

Using the sandwich rule (which i understand) find the limit of n!/nn

The attempt at a solution

To my knowledge n! is the fastest growing function you can have, so I immediately thought the function did not have a limit, however, the answer states the limit to be 1

I know that the bound on n! would be 1 =< n! < inf (?) but then dividing through by nn doesn't really help.

I am stuck :/

What am I doing wrong?
Isn't it quite intuitive that 0 is correct?
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement

Using the sandwich rule (which i understand) find the limit of n!/nn

The attempt at a solution

To my knowledge n! is the fastest growing function you can have
Then your knowlege is incorrect. n! is the product of n terms each less than or equal to n. [itex]n^n[/itex] is the product of n terms, each equal to n. For all n, [itex]n!< n^n[/itex].

, so I immediately thought the function did not have a limit, however, the answer states the limit to be 1

I know that the bound on n! would be 1 =< n! < inf (?) but then dividing through by nn doesn't really help.

I am stuck :/

What am I doing wrong?
 
  • #11
Ray Vickson
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Homework Helper
Dearly Missed
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Homework Statement

Using the sandwich rule (which i understand) find the limit of n!/nn

The attempt at a solution

To my knowledge n! is the fastest growing function you can have, so I immediately thought the function did not have a limit, however, the answer states the limit to be 1

I know that the bound on n! would be 1 =< n! < inf (?) but then dividing through by nn doesn't really help.

I am stuck :/

What am I doing wrong?

n! is NOT the fastest growing function you can have; there are infinitely many functions that grow much faster than n! (for example, (n!)^2 or exp(n!) or n^(n!) to name a few).

Anyway, do you know Stirling's formula? It says
[tex] n! \sim \sqrt{2 \pi}\: n^{n + 1/2} e^{-n}, [/tex] for large n, where "~" means "is asymptotic to", which in turn means that the ratio of the two sides has limit 1 as n → ∞. Thus, [tex] \frac{n!}{n^n} \sim \sqrt{2 \pi} \: \sqrt{n} e^{-n} \rightarrow 0[/tex] as [itex] n \rightarrow \infty.[/itex].

If you feel a bit suspicious about the use of "~" you can squeeze the ratio between two easily-analyzed expressions, because the original Stirling approximation can be turned into a rigorous pair of inequalities that are true for all n > 1:
[tex] \sqrt{2 \pi}\: n^{n + 1/2} e^{-n} < n! < \sqrt{2 \pi} \: n^{n + 1/2} e^{-n + (1/(12 n))} \:\:\forall n > 1. [/tex] For a simple proof, see,. eg., W. Feller, "Introduction to Probability Theory and its Applications", Volume I, Wiley (1968).

RGV
 
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