An interesting infinite product involving the factorial function with connection to the K and gamma function

Question

I have posted this question in StackExchange, but it didn't get any answers there. This question is important for my research. I got stuck on an infinite product which even WolframAlpha can't answer. Here's it: $$\prod_{n=2}^{\infty}\left(1-\frac{1}{n!}\right)$$

• This is surely convergent, many tests work. Wolfram Alpha couldn't evaluate it, but gave an approximate value of $$0.395338567367445566032356200431180613$$

• The decimal expansion is OEIS A282529, but the entry doesn't have much information. This constant is conjectured to be irrational, transcendental, and normal.

• This Math.SE question asks specifically for a closed form, but it has no answers, so it doesn't solve my question.

Here's the work I did: \begin{align} \prod_{n=2}^{\infty}\left(1-\frac{1}{n!}\right)&=\lim_{N\to\infty}\frac{\prod_{N\geq n\geq2}(n!-1)}{\prod_{N\geq n\geq2}n!}\\[6pt] &=\lim_{N\to\infty}\frac{\prod_{N\geq n\geq2}(n!-1)}{1\cdot1\cdot2\cdot1\cdot2\cdot3\cdots1\cdot2\cdot\cdots N}\\[6pt] &=\lim_{N\to\infty}\frac{\prod_{N\geq n\geq2}(n!-1)}{1^N2^{N-1}3^{N-2}\cdots(N-1)^2N^1} \end{align} Now I don't know how to proceed. L'Hopital's rule doesn't work, since the numerator isn't a function of $N$ (it is, but the product should be solved before differentiating).

How can I evaluate it? A link to an article containing information about the constant will also help. Any help would be appreciated.

Note: A closed form isn't necessary; converting the product into a sum or integral will also help. Some special function representations will also be good.
I realized that what I did was not useful. I did some research and found these facts:

• A representation of the Barnes-G function is $$G(N)=\frac{\Gamma(N)^{N-1}}{K(N)}$$ Where $K$ is the K-function.

• A representation of the K-function is $$K(z)=\mathrm{exp}[\zeta'(-1,z)-\zeta'(-1)]$$

Now I used the first point and simplified the product to $$\prod_{n=2}^{\infty}\left(1-\frac{1}{n!}\right)=\lim_{N\to\infty}\frac{K(N+2)}{\Gamma(N+2)^2}\prod_{k=2}^{N}\frac{k!-1}{(N+1)!}$$ How can this be simplified? Is there any suction related to this? The hard thing to evaluate is this: $$\prod_{k=2}^{N}(k!-1)$$ I looked up in this article but couldn't find a related function. Is there an article that discusses(or at least, mentions) this product?
Now my main question has become:

Simplify, give information about or represent in terms of special functions the product:$$\prod_{k=2}^{N}(k!-1)$$

Answer 1

I do not know if there is any closed form for this product, but you can rewrite it as follows. First, consider the logarithm of your product, so that you get: $$ L:=\log \left ( \prod_{n=2}^{\infty} (1-1/n!) \right) = \sum_{n=2}^{\infty} \log(1-1/n!)$$ Since $n \geq 2$, $\frac{1}{n!} < 1$, so we can use the Taylor series of the logarithm to obtain: $$ L=-\sum_{n=2}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k (n!) ^ k} $$ We can interchange the two series: $$ L= -\sum_{k=1}^{\infty} \frac{1}{k} \sum_{n=2}^{\infty} \frac{1}{(n!) ^ k} $$ For $k=1$, the inner sum is $e-2$. For $k=2$, according to wolfram alpha the inner sum is equal to $I_0(2) - 2$, where $I_0$ denotes the modified Bessel function of the first kind. For $k \geq 3$, wolfram alpha gives the result $_0 F _{k-1} (; 1, ..., 1; 1) - 2$ (the number of one's excluding the last one is $k-1$, of course). So define $a_k$ to be the sum of the $k$-th series above. Then: $$ L=-\sum_{k=1}^{\infty} \frac{a_k}{k} $$ Thus, your product is: $$ \prod_{n=2}^{\infty} (1-1/n!) = e^L $$ I do not think that there is a known closed form for $L$, but at least you can rewrite the product in terms of some known functions, as you requested.

Answer 2

I also don't think that there's a more closed form than what we have, but let it be known that if you go through the steps of rewriting outlined in the other two threads already linked (as well as here), you can get rid of the infinite product in favour of a more gaugable sum. Namely, the product equals

$$\frac{1}{2}\, -\, \sum_{n=3}^\infty\, \frac{1}{n!}\cdot\prod_{k=3}^{n-1}\left(1-\frac{1}{k!}\right)$$

or, if we rewrite the first three terms via Egyptian fractions,

$$\frac{1}{3} + \frac{1}{16} + \frac{1}{6912}\, -\, \frac{1}{2}\sum_{n=6}^\infty\, \frac{1}{n!}\cdot\prod_{k=3}^{n-1}\left(1-\frac{1}{k!}\right)$$

explicitly suggesting the range of your result $0.395338567\dots$

and where the summands are all dominated by $\frac{1}{n!}$.

Here's the routine:

1/2 - N[Sum[Product[1 - 1/k!, {k, 2, n - 1}] / n!, {n, 3, 100}], 50]