It is well known that $\displaystyle \lim_{n\to \infty}\sqrt[n]{n!}=\infty$, however, if we let $n\to 0$ we have a different result with a beautiful combination of $e$ and $\gamma$, that is

$$\lim_{n\to 0}\sqrt[n]{n!}= e^{-\gamma}\tag{1}\label{1}$$

To prove result \eqref{1}, we observe that the limit attains the form of $1^{\infty}$ so we can write it as $$\lim_{n\to 0} \exp\left(\frac{\ln\Gamma(n+1)}{n}\right)\underbrace{=}_{\text{L'Hopital's rule}}\lim_{n\to 0} e^{\Gamma'(n+1)}=e^{\psi_0(1)}= e^{-\gamma}$$

Now I wish to know the limit of the following multifactorial form for $k\in\mathbb {Z^+}$

$$\lim_{n\to 0}\sqrt[n]{n\smash[b]{\underbrace{!! !!\cdots !}_{k}}}={?}\tag{2}\label{2}\\$$

For $k=1$ we are done above and for $k=2$ we get the limit $\sqrt{2} e^{-\frac{\gamma}{2}}$. To prove this we use the double factorial argument (see equation (5)) \begin{align}\lim_{n\to 0}\sqrt[n]{n!!}&=\lim_{n\to 0} \left(2^{\frac{n}{2}+\frac{1-\cos(\pi n)}{4}}\pi^{\frac{\cos(\pi n)-1}{4}}\Gamma\left(1+\frac{n}{2}\right)\right)^{\frac{1}{n}}\\&=\sqrt{2}\lim_{n\to 0} \sqrt[n]{\Gamma\left(1+\frac{n}{2}\right)}\\&=\sqrt 2\exp\lim_{n\to 0}\left(2^{-1} \Gamma\left(1+\frac{n}{2}\right)\psi_0\left(1+\frac{n}{2}\right)\right)\tag{L'Hopital's rule}\\&=\sqrt{2}e^{-\frac{\gamma}{2}}\end{align}

since for $k=1,2$ we have evaluated the limit. How to evaluate the limit of equation \eqref{2} for all $k>2$?

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    $\begingroup$ Is the multifactorial here equivalent to $$ \frac{k^{\frac{n-1}{k}} \Gamma \left(\frac{n}{k}+1\right)}{\Gamma \left(1+\frac{1}{k}\right)} $$? $\endgroup$ – Benedict W. J. Irwin Aug 3 '20 at 13:43
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    $\begingroup$ A warning to others: In Mathematica $n!!!$ is not the triple factorial, it is equivalent to $(n!!)!$. $\endgroup$ – Benedict W. J. Irwin Aug 3 '20 at 14:36
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    $\begingroup$ @ Benedict, your multifactorial expression holds only for $k=1$ and for $k=2$ it holds only if $n$ is odd integer. For $k=3$ it doesn't hold. I agree with you regarding $n!!!$ since before making post I checked in WA. It interprets it's as $(n!!)!$. $\endgroup$ – Naren Aug 4 '20 at 5:57
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    $\begingroup$ It seems like the real question here is how to generalize multifactorials to complex arguments in a way that's akin to how the gamma function generalizes factorials. There may not be a straightforward answer, see math.stackexchange.com/questions/3191739/… and math.stackexchange.com/questions/291890/…. Certainly there's not a unique holomorphic or meromorphic interpolation, so it's conceivable it could end up depending on the gamma-like properties you want the function to satisfy. $\endgroup$ – Jason Aug 4 '20 at 7:05

The main task in this problem is to construct an analytic continuation of the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ over the real numbers. On the set of integers, we can take advantage of modulo arithmetic to arrive at $$n!^{(k)}_i=i\cdot k^{(n-i)/k}\frac{\Gamma(1+n/k)}{\Gamma(1+i/k)}$$ which is defined whenever $n\equiv i\pmod k$. Therefore, we need to find a function which interpolates the values of $$1,\frac{k^{-1/k}}{\Gamma(1+1/k)},\frac{2k^{-2/k}}{\Gamma(1+2/k)},\cdots,\frac{(k-1)k^{-(k-1)/k}}{\Gamma(1+(k-1)/k)}$$ so we consider the ansatz $$f(x)=\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{g(x)}$$ where $g(x)\equiv1$ when $x-i\equiv0\pmod k$ and is zero otherwise. This suggests the analytic function $$\frac{\sin(\pi(x-i))}{k\sin(\pi(x-i)/k)}$$ which is unfortunately negative when $(x-i)/k$ is an odd integer. We can circumvent this by introducing a factor of $\cos(\pi(x-i)/k)$ as its sign agrees with that of the above function when $x-i\equiv0\pmod k$. It follows that we can extend the multifactorial function to the reals through $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}\tag1$$ It is worth noting that this is not unique, as we can multiply by a factor of $\cos(j\pi(x-i)/k)$ for some integer $j$ — here we just took $j=1$. We can now determine the limit by considering each term separately \begin{align}\lim_{x\to0}x!^{(k)/x}&=k^{1/k}\lim_{x\to0}\exp\left(\frac{\log\Gamma(1+x/k)}x\right)\lim_{x\to0}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/kx}\\&=k^{1/k}e^{-\gamma/k}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{h(k)}\end{align} where \begin{align}h(k)&=\lim\limits_{x\to0}\frac{\sin(\pi(x-i))\cot(\pi(x-i)/k)}{kx}=-\cot\frac{\pi i}k\lim\limits_{x\to0}\frac{\sin\pi x\cos\pi i}{kx}\\&=-(-1)^i\frac\pi k\cot\frac{\pi i}k.\end{align} Using this definition of $g(x)$, the limit evaluates to $$\lim_{x\to0}x!^{(k)/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}.$$ Note that the multifactorial function in $(1)$ can be extended onto the complex plane. It is holomorphic everywhere except at negative integer multiples of $k$, similar to the gamma function.

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    $\begingroup$ It's late and I have insomnia but I had a realization that perhaps the notation for a multifactorial could be written as $$n \overset{k}{.}$$ since $k = 1$ looks like $$n\overset{1}{.} = n!$$ It's a sort of visual pun I guess. Surely I cannot be the first to have thought of this. $\endgroup$ – heropup Feb 17 at 11:54
  • $\begingroup$ I tried in Mathematica: Multifactorial[x_, k_] := k^(x/k)*Gamma[ 1 + x/k] Product[( (j*k^(-j/k))/ Gamma[1 + j/k])^(1/k*Sin[Pi (x - j)] Cot[Pi*(x - j)/k]), {j, 0, k - 1}]; Multifactorial[1, 1] give me: Indeterminate ? What is wrong ? $\endgroup$ – Mariusz Iwaniuk Feb 17 at 17:01
  • $\begingroup$ @MariuszIwaniuk Ah, I made a typo in that the product should go from $i=1$ not $i=0$. For those without Mathematica, here is a visualisation. $\endgroup$ – TheSimpliFire Feb 17 at 17:25

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