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