# $\infty$-multifactorial: $\displaystyle z!_{(\infty)}:=\lim_{\alpha\to\infty}z!_{(\alpha)}$

## Introduction

Let $$z!_{(\alpha)}$$ the $$\alpha$$-multifactorial.

$$z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}$$ Where $$C_{\alpha}(z)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\left\lfloor\frac{\alpha-1}{2}\right\rfloor}\cos\left(\frac{2k\pi}{\alpha}z\right)+\text{mod}(\alpha-1,2)\cos(\pi z)\right)$$

## My intent

I would like to calculate the limit of the multifactorial formula for $$\alpha$$ which approaches infinity.

Leaving aside the various steps that are long and irrelevant to the question

I arrived at this formula:

$$z!_{(\infty)}=\exp\left(\sum_{j=1}^{\infty}\text{sinc}(z-j)\ln(j)\right)$$

Where

• $$\text{sinc}(x):=\begin{cases}\dfrac{\sin(\pi x)}{\pi x}&\text{if }x\neq0\\1&\text{if }x=0\end{cases}$$

Unfortunately I can't move forward with this last sum :/, does anyone have any suggestions?
(It would also be acceptable to write it as an integral)

At the end of everything, a function with the following characteristics must emerge:

• $$n!_{(\infty)}=n$$ for $$n\in\mathbb{N}^{+}$$
• $$n!_{(\infty)}=1$$ for $$n=-1,-2,-3,...$$
• $$\displaystyle\prod_{i=0}^{\infty}(z-i)!_{(\infty)}=z!$$

This is the graph:

I'll also leave this link which leads to the desmos graph I made in case anyone wants to try their hand at trying to solve it.

## Update 1

The series is equal to $$z!_{(\infty)}=\exp\left(\frac{\sin(\pi z)}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{\ln(n)}{z-n}\right)\qquad \text{for }z\not\in\mathbb{N}^{+}$$

I take inspiration from an answer to make a clarification:

It would be useful to be able to use a result like this:

$$\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{z-k}f(k)=\pi\frac{f(z)}{\sin(\pi z)}$$

Applying this result with $$f(n)=\begin{cases}\ln(n)&\text{if }n>1\\0&\text{if }n\leq 1\end{cases}$$ the result would be: $$z!_{(\infty)}=\begin{cases}z&\text{if }z>1\\1&\text{if }z\leq 1\end{cases}$$ Unfortunately:

• In this case it is not applicable because one of the hypotheses requires that $$f(n)$$ is entire and the logarithm is not (and it can not be continued analytically to an entire function).
• $$z!_{(\infty)}$$ is a complex variable function, $$\mathbb{C}$$ is not ordered so it's not possible to say things like "$$z>1$$" or "$$z\leq 1$$"
• According to this solution for "$$z\leq 1$$" the series should ALWAYS be $$0$$.

## Update 2

$$x!_{(\infty)}=\exp\left(-\text{sinc}(x)\sum_{n=1}^{\infty}\eta'(n)x^n\right)$$

Where $$\eta'(s)$$ is the derivative of the Dirichlet Eta function.

• What is the multifactorial ? Where does it come from ? I guess it is some extension of the factorial but it would be helful to have a more "readable" definition. Closed formulas involving integers parts/modulos are rarely useful Commented Sep 19, 2023 at 7:29
• @Cactus You can read this answer: math.stackexchange.com/questions/1792900/…. It is possible to define $C_{\alpha}(x)$ even without using modules. As for the integer part being a limit, just consider $\alpha\mapsto 2\alpha$ Commented Sep 19, 2023 at 7:38
• @Cactus I edited the question and removed the dependence on $\alpha$ but I don't know how to do that last series. Commented Sep 19, 2023 at 9:20
• Interesting, it's a nice result if it works. Commented Sep 19, 2023 at 10:52
• It's unclear to me if $\sum_{j=1}^\infty\operatorname{sinc}(z-j)\ln(j)$ even converges. Commented Sep 19, 2023 at 10:55

This is a partial answer.

If I understand correctly from your comment, you extended the Wolfram alpha definition of $$z!! = z!_{(2)}$$ to all the multifactorials $$z!_{(\alpha)}$$ instead of using the Wikipedia definition, so the formula $$z! = \prod_{i = 0}^{\alpha - 1} (z - i)!_{(\alpha)},$$ holds for all $$z \in \mathbb{C}\backslash\mathbb{Z}_-$$, right ? We clearly have convergence of $$n!_{(\alpha)}$$ toward some ramp function when $$n$$ is an integer. Let $$z \in \mathbb{C}\backslash\mathbb{Z}$$.

Step 1 : simplifications and convergence of $$C_\alpha$$. First of all, $$\alpha^{\frac{z}{\alpha}}\Gamma\left(1 + \frac{z}{\alpha}\right) \rightarrow 1$$ so we can ignore this part. The only limit that remains to compute is the limit of, $$\prod_{j = 1}^{\alpha - 1} \left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_\alpha(z - j)} = \exp\left(\sum_{j = 1}^{\alpha - 1} C_\alpha(z - j)\ln\left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)\right).$$ Let us simplify the $$C_\alpha$$ now. For all real number $$x \in \mathbb{R}\backslash\mathbb{Z}$$ and all integer $$n$$, we have, \begin{align*} \sum_{k = 0}^n \cos(k\pi x) & = \Re\left(\sum_{k = 0}^n e^{ik\pi x}\right)\\ & = \Re\left(\frac{1 - e^{i(n + 1)\pi x}}{1 - e^{i\pi x}}\right)\\ & = \Re\left(\frac{e^{i\frac{(n + 1)\pi}{2}x}}{e^{i\frac{\pi}{2}x}}\frac{e^{-i\frac{(n + 1)\pi}{2}x} - e^{i\frac{(n + 1)\pi}{2}x}}{e^{-i\frac{\pi}{2}x} - e^{i\frac{\pi}{2}x}}\right)\\ & = \cos\left(\frac{n\pi}{2}x\right)\frac{\sin\left(\frac{(n + 1)\pi}{2}x\right)}{\sin\left(\frac{\pi}{2}x\right)}\\ & = (n + 1)\cos\left(\frac{n\pi}{2}x\right)\frac{\mathrm{sinc}\left(\frac{n + 1}{2}x\right)}{\mathrm{sinc}\left(\frac{1}{2}x\right)}. \end{align*} As $$\mathbb{R}\backslash\mathbb{Z}$$ is not discrete in $$\mathbb{C}$$ and by uniqueness of the analytic continuation, this equality remains true for any $$x \in \mathbb{C}\backslash 2\mathbb{Z} \cup \{0\}$$. In particular, if is true for any point in $$z + \mathbb{Z}$$.

If $$\alpha = 2\beta$$ is even, we have, \begin{align*} C_\alpha(z) & = \frac{1}{2\beta}\left(1 + 2\sum_{k = 0}^{\beta - 1} \cos\left(\frac{k\pi}{\beta}z\right) + \cos(\pi z)\right)\\ & = \frac{1}{2\beta}\left(1 + \cos(\pi z) + 2\beta\cos\left(\frac{(\beta - 1)\pi}{2\beta}z\right)\frac{\mathrm{sinc}\left(\frac{1}{2}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta}z\right)}\right)\\ & = \cos\left(\frac{(\beta - 1)\pi}{2\beta}z\right)\frac{\mathrm{sinc}\left(\frac{1}{2}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta}z\right)} + \mathrm{o}(1) \textrm{ when } \beta \rightarrow +\infty,\\ & \rightarrow \cos\left(\frac{\pi}{2}z\right)\mathrm{sinc}\left(\frac{1}{2}z\right)\\ & = \frac{2}{\pi z}\cos\left(\frac{\pi}{2}z\right)\sin\left(\frac{\pi}{2}z\right)\\ & = \mathrm{sinc}(z). \end{align*} And when $$\alpha = 2\beta - 1$$ is odd, similarly, \begin{align*} C_\alpha(z) & = \frac{1}{2\beta - 1}\left(1 + 2\sum_{k = 0}^{\beta - 1} \cos\left(\frac{2k\pi}{2\beta - 1}z\right)\right)\\ & = \frac{1}{2\beta - 1}\left(1 + 2\beta\cos\left(\frac{(\beta - 1)\pi}{2\beta - 1}z\right)\frac{\mathrm{sinc}\left(\frac{\beta}{2\beta - 1}z\right)}{\mathrm{sinc}\left(\frac{1}{2\beta - 1}z\right)}\right)\\ & \rightarrow \cos\left(\frac{\pi}{2}z\right)\mathrm{sinc}\left(\frac{1}{2}z\right)\\ & = \mathrm{sinc}(z). \end{align*}

Step 2 : convergence of the other part of the product. Fix some integer $$j$$. $$\Gamma(\varepsilon) \sim \frac{1}{\varepsilon}$$ when $$\varepsilon \rightarrow 0$$ thus, $$\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)} \sim \frac{j}{\alpha}\alpha^{\frac{\alpha - j}{\alpha}} = j\alpha^{-\frac{j}{\alpha}} \rightarrow j.$$ We deduce that, $$C_\alpha(z - j)\ln\left(\frac{\alpha^{\frac{\alpha - j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right) \rightarrow \mathrm{sinc}(z - j)\ln(j).$$

Step 3 : convergence of the final sum. The hard part is that the sum of the $$\mathrm{sinc}(z - j)\ln(j)$$ doesn't converge absolutely. However, we do have convergence. Indeed, we have for all $$j \in \mathbb{N}$$, $$\mathrm{sinc}(z - j) = \frac{\sin(\pi z - \pi j)}{\pi z - \pi j} = (-1)^j\frac{\sin(\pi z)}{\pi z}\frac{z}{z - j} = (-1)^j\mathrm{sinc}(z)\frac{z}{z - j}.$$ Therefore, $$\left|\mathrm{sinc}(z - j)\ln(j) - (-1)^j\mathrm{sinc}(z)\frac{\ln(j)z}{j}\right| = \left|\mathrm{sinc}(z)z\right|\left|\frac{1}{z - j} - \frac{1}{j}\right|\ln(j) \leqslant C\frac{\ln(j)}{j^2},$$ for some constant $$C$$ depending on $$z$$. The series of the $$\frac{\ln(j)}{j^2}$$ is absolutely convergent and the series of the $$(-1)^j\frac{\ln(j)}{j}$$ is convergent by an alternate sum argument. We deduce that the series of the $$\mathrm{sinc}(z - j)\ln(j)$$ converges. Moreover, we verify easily that the constant $$C$$ can be chosen locally uniformly with respect to $$z$$. It implies the local uniform convergence. In particular, $$z \mapsto \sum_{j \geqslant 1} \mathrm{sinc}(z - j)\ln(j)$$ is holomorphic on $$\mathbb{C}$$ i.e. it is an entire function.

Now, to conclude, we would need to bound the difference $$|C_\alpha(z - j)\ln(\cdots) - \mathrm{sinc}(z - j)\ln(j)|$$ by some $$K(j,\alpha) \geqslant 0$$ (ideally, $$K$$ would be defined in a neighborhood of each $$z \in \mathbb{C}\backslash\mathbb{Z}$$ to make sure we have local uniform convergence) such that $$\sum_{j = 1}^{\alpha - 1} K(j,\alpha) \rightarrow 0$$ when $$\alpha \rightarrow +\infty$$, but it seems very hard, especially for the $$j$$ close to $$\alpha$$.

• (+1) Congratulations, you wrote an extremely correct answer (you wrote the limit more efficiently than I did ahahah) Commented Sep 19, 2023 at 15:06
• Among other things, a "funny" thing is the fact that Wikipedia's definition doesn't even apply to integers. It defines $z!_{\left(\alpha\right)}=\alpha^{\frac{z-1}{\alpha}}\ \dfrac{\Gamma\left(\frac{z}{\alpha}+1\right)}{\Gamma\left(\frac{1}{\alpha}+1\right)}$. But if for example you try to calculate $8!_{(3)}:=8\cdot 5\cdot 2=80$ with that formula it comes approximately $58.320890...$ Commented Sep 20, 2023 at 12:42