0
$\begingroup$

If so, how can it be proven? (I have evaluated it up to $n=25$.)

If not, does there exist a $k\in\mathbb{R}$ such as that $n> a^k\Rightarrow n!>a^n$, with $n\in\mathbb{N},a\in\mathbb{R}$?


It is true, and to prove it, it suffices to show that $n!^2\geq n^n$ with induction.

For $n=1$ we have that $1\geq1$, which is true. Suppose $n!^2\geq n^n$. Then, $(n+1)!^2=(n+1)^2n!^2\geq (n+1)^2n^n$. We need to show that

$$\begin{align} &(n+1)^2n^n\geq (n+1)^{n+1}\Leftrightarrow\\ \Leftrightarrow &n^n \geq (n+1)^{n-1}\Leftrightarrow\\ \Leftrightarrow &\ln(n^n)\geq \ln((n+1)^{n-1})\Leftrightarrow\\ \Leftrightarrow &n\ln n\geq (n-1)\ln(n+1)\Leftrightarrow\\ \Leftrightarrow &\frac{n}{n-1} \geq \ln(n+1-n)=\ln1=0\quad, \end{align}$$

which is true!!

  • 1
    I imagine you can work out the answer. The key is [Stirling's approximation](http://en.wikipedia.org/wiki/Stirling%27s_approximation) If you do work it out, you should consider posting the answer yourself. (Yes, that is not only allowed, it is encouraged.)2011-11-05
  • 0
    It's true; it's a consequence of the fact that $n!^2\ge n^n$. As for how that's proved, perhaps I'll leave that unstated for now...2011-11-05
  • 0
    I proved $n!^2\geq n^n$ with induction.For $n=1$ we have that $1\geq1$ which is true. Suppose $n!^2\geq n^n$. Then, $(n+1)!^2=(n+1)^2n!^2\geq (n+1)^2n^n$. We need to show that $(n+1)^2n^n\geq (n+1)^{n+1}\Leftrightarrow (n+1)n^n\geq (n+1)^n\Leftrightarrow n^n \geq (n+1)^{n-1}$ $\Leftrightarrow \ln(n^n)\geq \ln((n+1)^{n-1})\Leftrightarrow n\ln n\geq \(n-1)\ln(n+1)$ $\Leftrightarrow \frac{n}{n-1} \geq \ln(n+1-n)=ln1=0$ which is true!! How could I use the Stirling's approximation to get a similar result??2011-11-05

1 Answers 1