Prove that, for all real number $r$, there is a constant $n(r)$ so that
\[n!>r^n\]
for all $n\geq r(n)$.
factorial vs powers
One one thing is neutral in the universe, that is $0$.
Re: factorial vs powers
If $n,n(r)\in\mathbb N$, I think it suffices to show that $\sqrt[n]{n!}$ is increasing, so eventually it will get past any constant $r$. Indeed we have $\sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}\Leftrightarrow n!^{n+1}<(n+1)!^n\Leftrightarrow n!<(n+1)^n$ which is obviously true for all $n$.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.