# Thread: finding the max and min value of n

1. ## finding the max and min value of n

Hi, what can be the limiting values of n for which $n!>a^n$ for any given a?
Thanks.

2. Or in other words for which minimal n the next inequality always follows for any a>0:
$n!^{1/n}>a$

just pick $n!^{1/n}=[a]+1$.

3. Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit

4. Originally Posted by pranay
Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit
I haven't worked this out completely, but I suspect you might find Stirling's approximation useful:
$\displaystyle n! \approx \sqrt{2 \pi n} \left ( \frac{n}{e} \right ) ^n$

That would give your problem the following form:
$\displaystyle \sqrt{2 \pi} n^{n + 1/2} > (ea)^n$

-Dan

5. Originally Posted by pranay
Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit
Well first thing first, what I wrote is the lowest n possible (there isn't a maximum cause you can always pick another one greater by adding 1 to the preceding number).

I am unaware of a closed analytical formula for what I wrote, I guess only Stirling approximation can work here (if you want to find some closed formula though still an approximation).