Show that for any $k>0$, there exists a $c>0$ such that $\frac{n}{e^{k n}} < \frac{c}{n^2}$ As the title says, I would like to show that for any constant $k > 0$, there exists a $c>0$ such that the inequality $\frac{n}{e^{k n}}< \frac{c}{n^2}$ holds for any $n\in\mathbb{N}\setminus\{0\}$. 
It is rather easy to see that such a $c$ exists when $k>1$, in which case $c=1$ would be sufficient. However, I am unsure how to start the problem for $k<1$. 
 A: Let $k > 0$.
Denote by $f : \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ be the function defined by $$f(x) = \frac{x^{3}}{e^{k x}}$$
Then $f$ is continuous on $\mathbb{R}_{\geq 0}$ and we have $$f(x) \underset{x \rightarrow +\infty}{\longrightarrow} 0$$
Therefore, $f$ is bounded above on $\mathbb{R}_{\geq 0}$.
Thus, for all integer $n \geq 1$, we have $$\frac{n}{e^{k n}} = \frac{f(n)}{n^{2}} < \frac{c}{n^{2}}$$ where $c = \sup\left\{f(x) : x \in \mathbb{R}_{\geq  0}\right\} +1$.
A: $\dfrac{n}{e^{kn}}
\lt \dfrac{c}{n^2}
\iff
n^3
\lt ce^{kn}
$.
Consider the more general problem
of showing that,
for any
$m, k > 0$
there is a 
$c = c(k, m)$
such that
$n^m
\lt ce^{kn}
$
for all integer
$n \ge 1$.
Taking the $k$-th root,
this is
$n^{m/k}
\lt c^{1/k}e^{n}
$,
which is the same problem
with $k=1$.
Therefore,
for any $r > 0$,
we want to find
$d > 0$
such that
$n^r \lt de^n$
for $n \ge 1$.
Here is a simple answer
that is far from the best
but still works:
Let $s = \lfloor r \rfloor +1$,
so
$n^r < n^s$
for all $n \ge 1$.
Then,
from the power series
for $e^x$,
$e^n
\gt \dfrac{n^s}{s!}$.
Therefore,
if
$n^s \le d\dfrac{n^s}{s!}$
we are done.
This is just
$d \gt s!$.
Going back to the
original statement,
$s = \lfloor m/k \rfloor +1$
and we want
$c^{1/k} > s!$
or
$c > (s!)^k$.
