# Homework Help: Need help with lim sqrt(n)/n! as n-> infinity

1. Nov 23, 2004

### twoflower

Hi,

could you help me a bit with this limit?

$$\lim_{n \rightarrow \infty} \sqrt[n]{n!}$$

Sure it should be more than

$$\lim_{n \rightarrow \infty} \sqrt[n]{n}$$

But, when I write it as

$$\lim_{n \rightarrow \infty} \sqrt[n]{n!} = \lim_{n \rightarrow \infty} \sqrt[n]{n} . \lim_{n \rightarrow \infty} \sqrt[n]{n-1} . \lim_{n \rightarrow \infty} \sqrt[n]{n-2} ... \lim_{n \rightarrow \infty} \sqrt[n]{1}$$

each term goes to 1, so I thought the limit could be 1, but that would be strange...

Thank you.

2. Nov 23, 2004

Hint: Use the corollary that (1+h)^ n > 1+ hn. Also you know that limit of nth root n is 1.

3. Nov 23, 2004

### arildno

Regard the following trick (regard even values of n):
$$n!=n^{n}\prod_{i=1}^{\frac{n}{2}}(1-\frac{i-1}{n})\prod_{j=1}^{\frac{n}{2}}(1-\frac{\frac{n}{2}+j-1}{n})$$
Now,
$$\prod_{i=1}^{\frac{n}{2}}(1-\frac{i-1}{n})\geq\prod_{i=1}^{\frac{n}{2}}(1-\frac{1}{2})=(\frac{1}{2})^{\frac{n}{2}}$$
And:
$$\prod_{j=1}^{\frac{n}{2}}(1-\frac{\frac{n}{2}+j-1}{n})\geq\frac{1}{n^{\frac{n}{2}}}$$

Hence,
$$n!\geq{n}^{n}(\frac{1}{2})^{\frac{n}{2}}\frac{1}{n^{\frac{n}{2}}}=(\frac{n}{2})^{\frac{n}{2}}$$

4. Nov 23, 2004

### twoflower

Thank you, but I can't see how could I use the inequality you suggested. Could you be more specific please?

5. Nov 23, 2004

### twoflower

So the thing you say is that

$$n! > \sqrt{ \frac{n^{n}}{2^{n}}}$$

I know this thing goes to infinity. And is THIS the reason, why also

$$\sqrt[n]{n!}$$

goes to infinity? But there is n-th root, why doesn't it make a difference?

6. Nov 23, 2004

### twoflower

Btw I can't imagine I would think up such a trick while writing a test

7. Nov 23, 2004

### BobG

8. Nov 24, 2004

### twoflower

9. Nov 24, 2004

### arildno

Of course, we have:
$$(n!)^{\frac{1}{n}}>((\frac{n}{2})^{\frac{n}{2}})^{\frac{1}{n}}=\sqrt{\frac{n}{2}}$$

10. Nov 24, 2004

### BobG

You open the lights and close lights, as well, right?

11. Nov 24, 2004

### twoflower

What's strange on writing ?

12. Nov 24, 2004

### rpc

because Americans "take" a test

13. Nov 24, 2004

### twoflower

Oh I see it now. It doesn't definitely mean I'm Canadian :) I'm just not that familiar with these phrases (although I know this one... )

14. Nov 24, 2004

### xepma

1st off: the reason why your method is wrong, is because you make the equality:
$$\lim_{n \rightarrow \infty} \sqrt[n]{n!} = \lim_{n \rightarrow \infty} \sqrt[n]{n}\lim_{n \rightarrow \infty} \sqrt[n]{n-1} \ldots$$

But when you take the limit of $n\rightarrow\infty$ the number of limits doesn't change as n increases. So you start of with n limits, multiplied with each other. Each individual limit has a limit of 1. But then you don't take into account that the number of limits multiplied together also changes, when n increases.

Anyways:
$$\lim_{n \rightarrow \infty} \sqrt[n]{n!} =$$
$$\lim_{n \rightarrow \infty} e^{\frac{1}{n}\ln{n!}}$$

Note again, that you can't write this as:
$$\lim_{n \rightarrow \infty} e^{\frac{1}{n}(\ln{n}+\ln{n-1}+\ln{n-2}+...)} =$$
and then take the limit of each logarithm seperately. Again, because the number of logarithms also changes with n.

Instead, you could for instance use the Stirling inequality:
$$\ln n! > n\ln{n} - n$$

$$\lim_{n \rightarrow \infty} e^{\frac{1}{n}\ln{n!}} >$$
$$\lim_{n \rightarrow \infty} e^{\frac{1}{n}(n\ln{n}-n)} = \infty$$