# Are infinite products commutative?

While reading a textbook, I came across the following proof (for integer partitions into odd parts and distinct parts):

The following steps can be justified by taking finite products and then passing to the limit:

\begin{align} \frac{1}{1-q}\frac{1}{1-q^3}\frac{1}{1-q^5} \ldots &= \frac{1}{1-q}\frac{1-q^2}{1-q^2}\frac{1}{1-q^3}\frac{1-q^4}{1-q^4}\frac{1}{1-q^5}\frac{1-q^6}{1-q^6}\ldots\\ &= \frac{1-q^2}{1-q}\frac{1-q^4}{1-q^2}\frac{1-q^6}{1-q^3}\ldots\\ &= (1+q)(1+q^2)(1+q^3)\ldots \end{align}

I'm sure that the above reasoning is fine, but I can't help but have a problem with the equality from the first line to the second.

To me, this raises the following question(s):

• Are infinite products commutative?
• If not, how might the above proof be justified using limits?
• You might want to consider the analogous theorem about sums, which is that you can rearrange the terms as much as you want as long as the sum converges absolutely. You can convert a convergent product to a (not necessarily absolutely) convergent sum by taking the logarithm. – MJD Aug 8 '13 at 7:21
• I see, converting products to sums via logarithms is quite a handy trick. – Sp3000 Aug 8 '13 at 12:10

## 1 Answer

One may go at this problem in two ways:

(i) Consider the occurring expressions $1\pm q^r$ as analytic functions of the complex variable $q$ with $|q|\ll1$. Using the fact that $$\bigl|{\rm Log} (1+z)\bigr|<{3\over2}|z|\qquad\bigl(|z|<{1\over2})$$ one can rewrite your identities as identities about absolutely convergent series. It follows that they are true as identities about certain analytic functions of $q$.

(ii) One may work in the realm ${\cal P}$ of formal power series (see, e.g., the first chapter of Henrici's Applied and computational complex analysis, I). Here ${1\over1-q}$ is an "abbreviation" for the series $1+q+q^2+\ldots$. Any such series is a bona fide element of ${\cal P}$, and an infinite sum or product of such series converges to a limit $a_0+a_1 q+a_2q^2+\ldots$, if each individual coefficient $a_r$ can be computed in only finitely many operations. To be exact: When $$a_0^{(n)}+a_1^{(n)} q+a_2^{(n)}q^2+a_3^{(n)}q^2+\ldots\qquad(n\geq1)$$ is a sequence of such series in ${\cal P}$ then this sequence converges to the series $$b_0+b_1q+b_2q^2+b_3q^3+\ldots$$ in ${\cal P}$ iff for each $r\geq0$ there is an $n_0=n_0(r)$ such that $$a_r^{(n)}=b_r\qquad\forall n\geq n_0\ .$$ It is easily seen that this is the case in your formulae, as later summands or factors only involve higher powers of $q$.

• Unfortunately, I don't know enough analysis yet to do the first option, but the second one seems viable. Are you saying that, because each individual coefficient $a_n$ can be computed via finitely many operations, the limit exists and is unique, meaning that "shifting down" the $1-q^{2k}$ factors does not change the limit? – Sp3000 Aug 8 '13 at 12:13
• @Sp3000: See my edit. – Christian Blatter Aug 8 '13 at 15:33