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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.

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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.

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Just as the terms in an infinite sum must converge to zero, so the terms in an infinite product must converge to +1. Which is not the case for any ordering of the rationals in an interval of positive length.

So your product is simply ill-defined, however you order the terms.

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