Solve for $k$ if $\beta^kk!\ge (1-\alpha)/\alpha$. Let $\beta$ be a constant and $\alpha\in (0,1]$. I want to show that for any $\alpha\in (0,1]$ (no matter how small) there exists $k\in\mathbb N$ such that $$\beta^kk!\ge (1-\alpha)/\alpha$$ 
I used that $\beta^kk!=(\beta^{-k}/k!)^{-1}$ and hence, since $x^k/k!\to 0$ as $k\to\infty$ for any $x\in \mathbb R$ (from the summation property of the exponential series), I obtained that such a $k$ always exists. 
My question is whether I can solve the above inequality for $k$ and derive a statement using for instance the big $\mathcal O$ notation, e.g., something like $k\in \mathcal O(1/\alpha)$. Any ideas? Thank you.
 A: From
$b^kk!\ge (1-a)/a
$
we get
$k\ln b +\ln(k!) \ge c$
where
$c = \ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$\ln(k!) > k\ln k - k$.
Then if
$k\ln b+k\ln k - k 
\ge c$,
$k$ is ok.
Write this as
$c 
\le k(\ln(b)-1) + k\ln(k)
=k(\ln(k)+\ln(b)-1)
=k(\ln(kb/e))
$
or
$cb/e
\le (kb/e)(\ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r \le x\ln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/\ln(r)$.
This becomes
$kb/e 
\approx \dfrac{cb/e}{\ln(cb/e)}
$
or
$k 
\approx \dfrac{c}{\ln(cb/e)}
$.
That's all.
A: If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $\beta^k\,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$k\sim e \beta\, e^{W(t)}-\frac 12=\frac{t e \beta}{W(t) }-\frac 12 \qquad \text{where}\qquad t=\frac{\log \left(\frac{A^2}{2 \pi  \beta }\right)}{2 e \beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t \ge-\frac 1e$.
