In the past I've known the post [1] from MathOverflow, and I am curious about if is known or you can to create an infinite product (converging) in a same way for an (arithmetic) function $f(n)$ as the basis of our factors and with exponents $B_n$ the Bernoulli numbers for $n\geq 1$. See this MathWorld if you need the definition of this sequence.
Question. Is feasible to define a convergent infinite product $$\prod_{n=1}^\infty \left(f(n)\right)^{B_n},$$ for a suitable $f(x)$, being $B_n$ the Bernoulli numbers? If you can not create such example and there are well known examples from the literature refer the literature and I try find it. Many thanks.
Only is required an example showing convergence, thus isn't required a closed-form. If you show a didactic way to find examples it is good, but isn't required.
I presume that I need to take logarithms and work with some asymptotic for the Bernoulli numbers. See the following codes using Wolfram Alpha online calculator if you want
prod n^(Bernoulli(n)), from n=1 to 14
prod n^(Bernoulli(n)), from n=1 to 16
prod n^(Bernoulli(n)), from n=1 to 18
thus I suspect that choose $f(n)=n$ is a bad way.
References:
[1] $\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$, from MathOverflow.
As stated in the comments $f(n)=1$ is convergent to $1$,
Also since $B_{2n} = \frac{(-1)^{n+1} 2 (2n)!}{ (2 \pi)^{2n}} \zeta(2n)$
It then follows from $\zeta \to 1$ $(n \to \infty)$ and Stirling's formula that
$|B_{2n}| \approx 4 \sqrt{n \pi} (\frac{n}{e \pi})^{2n} $,
So the family of all functions of the form $f(n) = n^{g(n)}$ such that $ g(n) << \frac{1}{ 4 \sqrt{n \pi} (\frac{n}{e \pi})^{2n} * n^{1+\epsilon} \ln^2 n} $ for all $\epsilon >0$.
Will be convergent, easily be showing that $\sum \limits_{ n =1}^{\infty} B_n \ln f(n) = \sum \limits_{ n =1}^{\infty} g(n) B_n \ln n \leq \sum \limits_{n=1}^{\infty} \frac{1}{n^{1+\epsilon} \ln^2 n} \ln n $ which is obviously convergent each to a different value.