Define the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ where the product extends to the least positive integer of $n$ modulo $k$. In this answer, I derived one of several analytic continuations of this function to the real numbers, which is as follows $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}.$$ The limit of original interest $$F(k)=\lim_{x\to0}\,(x!^{(k)})^{1/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}$$ is a straightforward consequence of the above result. Out of curiosity I plotted $F(k)$ and it appears that the double limit $$m=\lim_{k\to\infty}\lim_{x\to0}\,(x!^{(k)})^{1/x}$$ exists, converging rapidly to around $0.852$ when $k\in(s-1/2,s+1/2)$ for all positive odd integers $s$, but converging much slower to the same value when $k\in(s+1/2,s+3/2)$.

We can rewrite the limit as \begin{align}m&=\lim_{k\to\infty}\exp\left(\frac{-\gamma+\log k+\pi\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)}k\right)\\&=\lim_{k\to\infty}\exp\left(\frac\pi k\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)\\&=\exp\left(\lim_{k\to\infty}\frac\pi k\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\frac{k^{i/k}\Gamma(i/k)}k\right)\end{align} using L'Hopital on $(-\gamma+\log k)/k$. The term inside the exponential looks very much like a Riemann sum but I'm not sure where to go after that. It seems that none of the terms in the logarithm can be split additively to evaluate the limit as each component by itself is divergent.

Is there a closed form for this double limit?


1 Answer 1



$$\boxed{\;\ln m=-\gamma\ln2+\frac{\ln^22}{2}\;}$$

Explanation (sketch of the proof): for $i,k\to\infty$ such that $i/k=x$ with finite $x>0$ the expression under the sum $\sum_{i=1}^{k-1}$ tends to $0$ and therefore this range of $i$ does not contribute to the limit. We may therefore restrict the summation to the values with $i/k\ll 1$. Expand everything in $i/k$: $$\cot\frac{\pi i }{k}\to \frac{k}{\pi i },\qquad \ln\frac{\Gamma(i/k)}{k}\to -\ln i,\qquad \ln k^{i/k}=i\frac{\ln k}{k}.$$ The third piece should not contribute to the limit $k\to\infty$ because of the cutting factor $\frac{\ln k}{k}$. The first two produce the following expression in the exponential: $$\sum_{i=1}^\infty (-1)^{i+1}\frac{\ln i}{i}=-\gamma\ln2+\frac{\ln^22}{2}.$$ There admittedly remain some details to be filled for the above to become a rigorous proof, however I leave it to whoever is interested.

  • 9
    $\begingroup$ (+1) Beautiful, $\lim\limits_{k\to\infty}\lim\limits_{n\to 0} \sqrt[n]{n\underbrace{!!!!\cdots!}_{k\,\text{times}}}=2^{\log\sqrt2-\gamma}$. For those unfamiliar with the identity in the last equation see The sum of $(-1)^n\frac{\ln n}n$. $\endgroup$
    – TheSimpliFire
    Feb 21, 2021 at 13:50

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