I'd like to know if the following statement is a valid one. Specifically, I hope to use it in applying comparison tests for convergence/divergence of certain series.
For sufficiently large n,
$n, n^2, n^3, n^4, ... < 2^n, 3^n, 4^n, ... < n!$
The statement wasn't in my text, and I couldn't find it anywhere -- but I tried to recall it from back when I took Calculus II myself.
Is the above statement correct ? Further, can the above statement be expanded upon to say
For sufficiently large n,
$n^c < a^n < n!$ ,
where $c$ is any positive real number and $a$ is any real number greater than $1$ ?
Thanks !
Proofs are relatively easy.
$n^c \lt a^n$ can be gotten by taking log. $clog(n)\lt nlog(a)$ or $\frac{c}{log(a)}\lt \frac{n}{log(n)}$ for large enough $n$.
Using Stirling formila $n!\approx \sqrt{2\pi n}(\frac{n}{e})^n$ allows $a^n\lt n!$ to be $1\lt \sqrt{2\pi n}(\frac{n}{ae})^n$, as long as $n\ge ae$.
The expansion is more general (as I interpret it) The first inequality looks like $n^k\lt k^n$ for fixed $k$ and large $n$.