Why Does n!/n^n Approach Zero?

[Tomp]

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?

 

[Tomp]

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

 

[DryRun]

What limits did you compare it to, for the Sandwich theorem?

 

[Infinitum]

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?

 

[Tomp]

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

 

[Tomp]

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...

 

[Infinitum]

apologies as n approaches infinity

Aye, 0 is correct, then.

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

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...

 

[DryRun]

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.

 

[dimension10]

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?

 

[HallsofIvy]

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. n^n is the product of n terms, each equal to n. For all n, n!&lt; n^n.

, 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?

 

[Ray Vickson]

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
n! \sim \sqrt{2 \pi}\: n^{n + 1/2} e^{-n}, 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, \frac{n!}{n^n} \sim \sqrt{2 \pi} \: \sqrt{n} e^{-n} \rightarrow 0 as n \rightarrow \infty..

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:
\sqrt{2 \pi}\: n^{n + 1/2} e^{-n} &lt; n! &lt; \sqrt{2 \pi} \: n^{n + 1/2} e^{-n + (1/(12 n))} \:\:\forall n &gt; 1. For a simple proof, see,. eg., W. Feller, "Introduction to Probability Theory and its Applications", Volume I, Wiley (1968).

RGV