Can the infinite product of the powers of a transcendental number ever be transcendental?

Question

I plugged the following product into a calculator: $$\prod_{n=1}^\infty e^{\frac{1}{n^2}}$$ and got a result of roughly 5.1806683. I would like to say that this result is transcendental, as its only products are the positive rational powers of a transcendental number, but I'm a bit afraid of something like this happening: $$\sum_{n=0}^\infty \frac{(-1)^n}{\pi(2n+1)} = 1/4$$ Or does stuff like that only happen with summation?

Answer 1

First of all $\displaystyle \prod_{n=1}^{\infty} e^{\frac{1}{n^{2}}} = e ^{\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}} = e^{\frac{\pi^{2}}{6}}$. Now, $e^{\pi}$ is a transcendental number. This follows from the solution to Hilbert’s 7th problem, proved by Gelfond and Schneider in the 1930’s. But unfortunately, $e^{\pi^{2}}$ are not known to be rational, algebraically irrational or transcendental.

And

$ \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \pi/4$, for proof of this result you can see this answer . It is a transcendental number as $\pi$ is transcendental.