# Does order matter for the convergence of infinite products

Similar to infinite sums, does order matter in the convergence of infinite products? More specifically, I'm interested in the product of all rational numbers in the interval $(0,a]$.

For example, let $a=3$. I assert that the product converges to 0. Since all rationals in the interval $[\frac{1}{3},1)$ have a unique inverse in the interval $(1,3]$, we are left with the infinite product of all rationals in the interval $(0,\frac{1}{3})$, which converges to 0.

Now, consider the following product:

$$\prod _{i=0}^{n-1}{\frac{3n-3i}{n}}=3\cdot\frac{3n-3}{n}\cdot\frac{3n-6}{n}\cdot \ldots \cdot \frac{6}{n}\cdot\frac{3}{n}$$

As a quick example, when $n=18=3\cdot3!$, we get (in simplest form): $$3\cdot\frac{17}{6}\cdot\frac{8}{3}\cdot\frac{5}{2}\cdot\frac{7}{3}\cdot\frac{13}{6}\cdot2\cdot\frac{11}{6}\cdot\frac{5}{3}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{7}{6}\cdot1\cdot\frac{5}{6}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}$$

The important thing to see here is that we have all rational numbers from 0 to 3 with denominators of 1,2, and 3 in our product (as well as some other rationals). In fact, if we let $n=3k!$ for some $k\in\mathbb{N}$, we will get all rational numbers from 0 to 3 with denominators of k. Therefore, if we let $n\to\infty$, then (unless I'm mistaken) we should get the product of all rationals from 0 to 3. This, however, diverges to infinity:

$$\lim_{n \to \infty}\prod _{i=0}^{n-1}{\frac{3n-3i}{n}}=\lim_{n \to \infty}\frac{3^n}{n^n} n!=\infty$$

With that being said, why does one of the products converge and the other diverge? My two guesses there's some condition on infinite products for convergence, or I am missing something in the second product.

An infinte product $\prod_{n=1}^\infty a_n$ with $a_n>0$ is said to converge if and only if $\sum_{n=1}^\infty \ln a_n$ converges. (And a product inovlving negative or zero factors is allowed only if there are at most finitely many of them) Note that this is more restrictive than saying that $\lim_{m\to\infty}\prod_{n=1}^m a_n$ converges. Just like the convergence and limit of $\sum_{n=1}^\infty \ln a_n$ may depend on the summation order, the same applies to the infinite product. Only if the series converges absolutely, i.e. $\sum_{n=1}^\infty |\ln a_n|<\infty$, the order of summands (and hence also of factors) is irrelevant.
We can translate the condition for absolute convergence to products: If we replace all factors $<1$ by their reciprocals and then the sequence of partial products still converges, then the original product does not depend on the order of factors. Since $|\ln x|$ becomes arbitrarily big when $x$ approaches $0$, your product of rationals in $(0,a]$ indeed depends on factor order.