Infinite product of transcendental numbers approaches 1

Question

I am seeking infinite formulas connect transcendentals and rationals. We know $$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$$

as an example of infinite sum of rational numbers which approaches the transcendental e. Is there an infinite product exists as a kind of dual counterpart of the above Euler's formula? i.e. which transcendentals $\{x_n\}$ satisfied

$$1 = \prod_{n = 0}^\infty x_n$$

The beauty the better, just like Euler's version.

Answer 1

The gamma function $\Gamma(x)$ satisfies the identity below, where $\gamma$ denotes the Euler-Mascheroni constant. The identity is evidently a result of the Weierstrass factorization theorem (see the examples section).

$$\frac{1}{\Gamma(x)} = xe^{\gamma x} \prod_{n=1}^\infty \left(1 + \frac x n \right) e^{-x/n}$$

Take $x=1$. Then, since $\Gamma(x) = (x-1)!$ for positive integers $x$, and $0!=1$, we get

$$1 = e^\gamma \prod_{n=1}^\infty \left(1 + \frac 1 n \right) \frac{1}{\sqrt[n]e}$$

We don't know if $\gamma$ itself is transcendental or not (that question is still open), but the $n^{th}$ roots of $e$ are transcendental as a corollary of Lindemann-Weierstrass.

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Some potential flaws with this:

• This isn't a product from $0$ to $\infty$, but a shift of index fixes that
• $e^\gamma$ isn't known to be transcendental, and it might somehow negate that of the factors $x_n = (1+1/n)e^{-1/n}$

Granted, I think there might be merit in this example - it certainly "feels" likely that the factors are transcendental, even if we can't quite prove that yet. There might also be merit in exploring Weierstrass factorization further for something that can be more surely seen to have transcendental factors. Perhaps with different functions, for instance?