Homework Help: Limit of a sequence n!/n^n

1. Jun 1, 2012

Tomp

The problem statement, all variables and given/known data

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?

2. Jun 1, 2012

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

3. Jun 1, 2012

sharks

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

4. Jun 1, 2012

Infinitum

Just asking what the limit of f(x) is, gives no meaning. You need to specify to what value n approaches?

5. Jun 1, 2012

Tomp

apologies as n approaches infinity

6. Jun 1, 2012

Tomp

i said the lower limit for n! was 1 and the upper limit was infinity....

7. Jun 1, 2012

Infinitum

Aye, 0 is correct, then.

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

From this reasoning(incorrect), I don't see how you got 0 as your answer though....

8. Jun 1, 2012

sharks

And how did you end up with those limits? Just curious.

9. Jun 1, 2012

dimension10

Isn't it quite intuitive that 0 is correct?

10. Jun 1, 2012

HallsofIvy

Then your knowlege 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, $n!< n^n$.

11. Jun 1, 2012

Ray Vickson

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

RGV

Last edited: Jun 1, 2012