What is the closed form approximation of the asymptotic growth rate of the superfactorial function?

Question

The asymptotic growth rate of the hyperfactorial function (defined to be: $H(n)=\prod^n_{k=1}k^k$) is apparently (approximately) equal to: $$ H(n) \sim An^{(6n^2 + 6n + 1)/12} e^{ - n^2 /4} $$ I'm curious as to how this result is obtained, and am also interested in the asymptotic growth rate of the superfactorial function as defined by Sloane and Plouffe (1995) : $$sf(x)= \prod^x_{k=1}k!$$

Answer 1

To answer your first question, we will make use of the followig property of the logarithm

$$ \log(\prod_k f(k) )=\sum_k\log(f(k)) $$

and therefore (using $k^k=e^{\log(k)k}$)

$$ L(n)\equiv\log(H(n))=\sum_{k=1}^n\log(k^k)=\sum_{k=1}^n\log(k)k $$

To make further progress we apply the Euler Maclaurin formula

$$ L(n)\approx \int_1^n x \log(x)dx+\frac{\log(n)n}{2}=\frac{n^2\log(n)}{2}+\frac{1}{4}-\frac{n^2}{4}+\frac{\log(n)n}{2} $$

and therefore

$$ H(n)\approx e^{\frac{n^2\log(n)}{2}+\frac{\log(n)n}{2}+\frac{1}{4}-\frac{n^2}{4}}=n^{n^2/2+n/2}e^{1/4-n^2/4} $$

Wait a second! This not quite correct because we are of by mulitplicative factor of $n^{1/12}$. What have we done wrong? The answer is, that we reexponiated to early because this is only allowed if the dismissed factors in the sum go to zero for large $n$. Luckily we can cure this by looking at the remainder term of the Euler Maclaurin expression

$$ R(n)=\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}(f^{(2k-1)}(n)-f^{(2k-1)}(1)) $$

with $f(x)=\log(x)x$ .Taking derivatives we see that $f'(n)=\log(n)+1$ and $f'''(n)=-\frac{1}{n^2}$. Higher derivatives vanish even faster as $n\rightarrow \infty$, so what we have missed above is the first term of the remainder together with the complete $n$ independent (denoted by $c$) part. Using $B_2=1/6$ this evaluates to

$$ \frac{1}{12}\log(n)+c $$

and therefore

$$ H(n)\approx e^{\frac{n^2\log(n)}{2}+\frac{\log(n)n}{2}+ \frac{\log(n)}{12}+\frac{4c+1}{4}-\frac{n^2}{4}}=n^{n^2/2+n/2+1/12}e^{(4c+1)/4-n^2/4} $$

which is the claimed result, if one identifies $A=e^{(4c+1)/4}$

Edit:

I have the feeling that this technique might be also capable to give the superfactorial asymptotics but i'm too busy to try it at the moment

Appendix

Ok i can't resist to give at least a sketch how one obtains the result stated by @Gerry Myerson

The key idea is the same as above, $$sf(n)=e^{l(n)}$$ with

$$ l(n)=\sum_{k=1}^n\log(k!) $$

and now we can again apply Euler MacLaurin.

The hard part here is the integral which gives the leading order of the sum

$$ \int_0^n \log(\Gamma(x))dx $$

Luckily there is a result due to Adamchik relating this integral to the Barnes $G$ function which is well known (see link above)

$$ \int_0^n \log(\Gamma(x))dx=-\left(\frac{n(1-n)}{2}+\frac{n}{2}\log(2 \pi)+\log(\Gamma(n))n\right)+\log(G(1+z)) $$

Putting this together with Stirlings formula for the Gamma function (and maybe the digamma function if we need higher order terms in the Euler Maclaurin formula), we should be able to arrive at the announced result

Answer 2

The superfactorials are discussed at https://oeis.org/A000178 where the asymptotic $$a(n) \sim \exp(\zeta'(-1)-3/4-3/4n^2-3/2n)(2\pi)^{1/2+1/2n}(n+1)^ {1/2n^2+n+5/12} $$ is given.

The hyperfactorials are discussed at https://oeis.org/A002109 and maybe one of the references given there will lead you to a proof.