# 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

Staff Emeritus
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