## Tuesday, September 15, 2015

### Prove $\large 1000!^{\frac{1}{1000}}>999!^{\frac{1}{999}}$.

Prove $\large 1000!^{\frac{1}{1000}}>999!^{\frac{1}{999}}$.

This inequality would be easy to prove if one uses the more advance knowledge, like Stirling's formula where it states when $n\rightarrow \infty$ then we have $n!\approx \dfrac{n^n}{e^n}\sqrt{2\pi n}$.

But, elementary method works well too in this problem.

Below is my solution that I want to share:

It is not hard to see that

$2^2\gt 2!$, $3^3\gt 3!$, and so on and so forth...

Therefore we have

$n^n\gt n!$ holds for all positive integer $n\ge 2$

$\dfrac{1}{n!}\gt \dfrac{1}{n^n}$

$\dfrac{n!^n}{n!}\gt \dfrac{n!^n}{n^n}$

$\dfrac{n!^n}{n!}\gt \dfrac{(n(n-1)!)^n}{n^n}$

$\dfrac{n!^n}{n!}\gt \dfrac{n^n(n-1)!^n}{n^n}$

$\dfrac{n!^n}{n!}\gt \dfrac{\cancel{n^n}(n-1)!^n}{\cancel{n^n}}$

$n!^{n-1}\gt (n-1)!^n$

$n!^{\frac{1}{n}}\gt (n-1)!^{\frac{1}{n-1}}$

Therefore, we have proved the more general case and if we let $n=1000$, we have

$\large 1000!^{\frac{1}{1000}}\gt (999)!^{\frac{1}{999}}$

and we're hence done.