Limit of a sequence n/n^n

In summary: I don't think it would give you a definitive answer, as it's only an approximation. I don't think it would give you a definitive answer, as it's only an approximation.
  • #1
Tomp
27
0
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?
 
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  • #2
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
What limits did you compare it to, for the Sandwich theorem?
 
  • #4
Tomp said:
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
Infinitum said:
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
sharks said:
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
Tomp said:
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
Tomp said:
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
Tomp said:
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
Tomp said:
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 knowledge 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
Tomp said:
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
 
Last edited:

What is the limit of the sequence n/n^n?

The limit of the sequence n/n^n is 0. This can be shown mathematically by taking the limit as n approaches infinity and using L'Hospital's rule.

How do you find the limit of n/n^n?

To find the limit of n/n^n, you can rewrite the expression as 1/n^(n-1) and then use L'Hospital's rule to take the limit as n approaches infinity.

Is the limit of n/n^n always 0?

Yes, the limit of n/n^n is always 0. This can be proven mathematically by using L'Hospital's rule or by considering the growth rate of the numerator and denominator as n approaches infinity.

What is the significance of the limit of n/n^n?

The limit of n/n^n is significant in calculus and analysis because it represents the behavior of a sequence as n approaches infinity. It also has applications in the study of series and convergence.

Can the limit of n/n^n be calculated for non-positive values of n?

No, the limit of n/n^n is undefined for non-positive values of n. This is because raising any number to a negative power results in a fraction with a denominator of 0, which is undefined.

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