Limit

[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.

[courtrigrad]

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

[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}}

This should help you..

[twoflower]

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

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

[twoflower]

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

This should help you..

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?

[twoflower]

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})


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

[BobG]

Btw I can't imagine I would think up such a trick while writing a test
:mad:
You must be Canadian.

[twoflower]

You must be Canadian.

Why should I be Canadian?

[arildno]

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

[BobG]

Btw I can't imagine I would think up such a trick while writing a test
:mad:
You speak Canadian (see bold).

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

[twoflower]

You speak Canadian (see bold).

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

What's strange on writing ?

[rpc]

What's strange on writing ?


because Americans "take" a test

[twoflower]

because Americans "take" a test

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... :approve: )

[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