Find a "nice" expression for the smallest j such that (2j)! > r, for a positive real number r?

Question

I am in the middle of a fairly complex limit proof, and am stuck at the point where I need to find a working formula/approximation in terms of r for the smallest integer argument j with (2j)! > r. Can anyone point me in the right direction to solve this issue?

Answer 1

Is Stirling's formula sufficiently accurate for your purposes? It bounds:

$$ \sqrt{2\pi n} \left( \frac n e \right)^n \leq n! \leq \frac e{\sqrt{2\pi}} \sqrt{2\pi n} \left( \frac n e \right)^n $$

By using the version with $n$ replaced by $n+1$, you can get the slightly better approximation

$$ \sqrt{\frac{2\pi}{n+1}} \left(\frac{n+1}e\right)^{n+1} \leq n! $$

In any case, this tells you that if you want $n=2j$ such that $n! > r$, it suffices to make sure $\sqrt{2\pi n} \left( \frac n e \right)^n > r$, or equivalently

$$ n \log n - n + \frac12 \log(2\pi n) > \log r $$

And on the other hand, the $n$ making $n! > r$ will satisfy

$$ n \log n - n + \frac12 \log(2\pi n) + \frac12 \log(2\pi) - 1 > \log r $$

So this bounds your sought-after $n$.

Even rounding up to $n \log n$ doesn't give an easy formula — I don't know any better way to describe the inverse function to $n \mapsto n \log n$ than as "the inverse function to $n \mapsto n \log n$".

If the usual Stirling's formula is not sufficiently accurate for your application, perhaps the next few asymptotics (also listed op. cit.) are enough.