Can the product of infinitely many elements from $\mathbb Q$ be irrational? I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. 
Thank you for your answers.
 A: Consider the Riemann-Zeta Function:
$$
\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}}.
$$
For $s=2$, the infinite sum on the left is $\pi^{2}/6$, which is irrational. Thus, $\pi^{2}/6$ is an infinite product of rationals. 
A: There is a simple way to obtain any irrational number as an infinite product:


*

*take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);

*form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.
$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$
A: Yes!  
$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$
A: Too big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:
\begin{align*}\cfrac{\pi}{2}&=\cfrac{2}1 \cfrac 23 \cdot \cfrac 43  \cfrac 45 \cdot\cfrac 65 \cfrac 67\cdot \ldots\\ &= \cfrac{2^2}{2^2-1}\cdot \cfrac{4^2}{4^2-1}\cdot \cfrac{6^2}{6^2-1}\ldots\\ \end{align*}
is an infinite product with partials starting at $\frac43$ and increasing towards $\frac\pi 2$ (every factor is greater than $1$), whereas the seemingly identical
\begin{align*}0&=\cfrac{2}3 \cfrac 23 \cdot \cfrac 45\cfrac 45\cdot\cfrac 67 \cfrac 67 \cdot\ldots\end{align*}
starts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left.
A: Yes, it can.
Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and
$$
  b_n = \frac{a_n}{a_{n-1}}
$$
for $n > 1$.
We then have that
$$
  b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.
$$
We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.
A: Yes, every irrational number is an infinite product of rationals.
We can write an infinite sum of rationals as an infinite product of rationals.
$$\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}$$ 
For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$
$$ 1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac     {1.414}{1.41}\times .....$$
