Irrationality of a number represented by an infinite product, whose partial products are irrational

Question

It is known that infinite sequences of irrational numbers can converge to rational numbers. For instance, the sequence: $$ \{x_n=\sqrt[n]{n}\}_{n=1}^\infty $$ equals $1$ when $n\to\infty$, however it is irrational for all $n>1$.

But what about infinite products? For instance, let the function: $$ f=\prod_{n=1}^{\infty}\sqrt[b]{a}, $$

for $b\in\mathbb{N}$, $b>1$ and $a$ irrational such that $f$ converges to the real number $k$. We immediately see that all partial products are irrational. Can we then also say that $k$ is irrational? If not, why?

Answer 1

Let $a=\sqrt{2}$ for all $n$ and $b_n =2$ for $n =1,2,3$ and $b_n=2^{n-2}$ for $n \ge 3$

Then you have $b_n$ as the sequence $2,2,2,4,8,16,\ldots$ and $\sum \frac{1}{b_n}=2$

So $$\prod_{n=1}^{\infty}\sqrt[b_n^{\,}]{a} = a^{\sum \frac{1}{b_n^{\,}}}=\sqrt{2}^2=2$$

which is an integer limit despite all the $\sqrt[b_n^{\,}]{a}$ being irrational

Answer 2

Let $p$ be a prime number and $n \in \Bbb N$. Clearly, $p^{1/n}$ is irrational.

Now, consider a sequence $(n_i)_{i \in \Bbb N}$ such that $\sum_{i=1}^\infty \frac{1}{n_i}=1$.

Observe that $\prod_{i=1}^\infty p^{1/n_i} =p$.