Solve for $k$ if $\beta^kk!\ge (1-\alpha)/\alpha$.

Question

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.

Answer 1

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.

Answer 2

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 $t \ge-\frac 1e$.