# When does $\prod_{k=1}^\infty F_k(x)$ converge?

I have a quick question about multiplying formal series.

For a little notation, I write $\langle x^n\rangle F(x)$ for the $n$-th coefficient in a formal series $F(x)$. Also, a sequence $\{F_k(x)\}$ in the formal power series ring $R[[X]]$ converges to $G(x)$ if the sequence $\{\langle x^n\rangle F_k(x)\}$ converges to $\langle x^n\rangle G(x)$, in the discrete topology on the ring $R$.

I few days ago I saw that $\sum_{k=1}^\infty F_k(x)$ converges iff $\{F_k(x)\}$ converges to $0$.

When can one conclude that $\prod_{k=1}^\infty F_k(x)$ converges? My guess is that it happens iff $\{F_k(x)\}$ converges to $1$. Is this indeed the case, or does extra precaution need to be made when multiplying?

(P.S. In light of Riemann's rearrangement theorem for convergent series, is this product affected in any way by rearranging the terms? Just an extra curiosity, if there's time to answer.)

Thank you,

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Your guess is correct in the case that the constant term of every $F_i$ is $1$, a condition that is true in practically every application of infinite products of formal power series. I'll cite R. Stanley's Enumerative combinatorics I on this:

1.1.8 Proposition. The infinite series $\sum_{j\geq0}F_j(x)$ converges if and only if $\lim_{j\to\infty}\deg F_j(x)=\infty$.

1.1.9 Proposition. The infinite product $\prod_{j\geq0}(1+F_j(x))$, where $F_j(0)=0$, converges if and only if $\lim_{j\to\infty}\deg F_j(x)=\infty$.

Note that Stanley, rather dubiously, calls degree of a (non-zero) power series what is actually its valuation: the smallest index of a non-zero coefficient. (His terminology gets prolematic for power series that happen to be polynomials). The limits being $\infty$ therefore means the same as the limits of the power series themselves being $0$. The point I cited this is that Stanley, who uses power series and infintie products all of the time, doesn't even consider it necessary to define infinite products at all unless the constant terms of all factors are $1$.

If the constant terms are all $1$, then it is clear that to have convergence, what remains of the factors without this constant term has to tend to $0$: multiplying a power series with constant term $1$ with another one that has a non-zero coefficient $c_n$ of $X^n$ will change the coefficient of $X^n$ of the former by adding $c_n$; clearly for any fixed $n$ this can happen only finitely often if the coefficient of $X^n$ is to stabilise ultimately.

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Thanks for this answer! – Adelaide Dokras Mar 4 '12 at 9:59

As I understand your definition of convergence a sequence of functions $F_k(x)$ converges to $G(x)$ if and only if for any $N$ you can find an $M$ such that $$n>M \ \Rightarrow \ F_n(x)-G(x) \in x^N R[[x]]$$ In light of this it seems that the constant sequence $F_k(x)=x$ will satisfy that $\prod_{k=0}^{\infty} F_k(x)$ will converge to $0$. To kill such trivial examples (the sequence $F_k(x)=0$ would be another), we can add the assumption $F_k(0) \neq 0$ for all $k$. In addition assume that the limit function $G$ satisfies $G\neq 0$. In this case the assumption $\prod_{k=0}^{\infty}F_k(x) \to G$ implies $$F_{M+1}G- G \in x^NR[[x]]$$
Since $G\neq 0$ there exists some $n_0$ such that $$G(x) = a_{n_0}x^{n_0}+ \dots$$ hence $$F_{M+1}(x) -1 \in x^{N-n_0}R[[x]]$$ hence $F_{k}$ does indeed converge to $1$.

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Actually the assumption $F_k(0) \neq 0$ for all $k$ is better replaced by $\prod_{i<k}F_i(0)\neq 0$ for all $k$ (just in case $R$ has zero divisors: to avoid the same kind of difficulty indirectly as $F_k(0)=0$ causes directly), and then the constant term will stabilise to a non-zero value, and you will get $G(0)\neq0$ for free. However zero divisors are still a pain: your conclusion that $F_{M+1}-1 \in x^{N-n_0}R[[x]]$ depends on simplifying by $a_{n_0}$ (which is in fact $a_0$), and if it is not a regular element of $R$, the conclusion is not warranted. – Marc van Leeuwen Feb 10 '12 at 18:40
Good point about the zero divisors, I didn' think about that. That seems to pretty much sink my argument in that case. – testcase Feb 10 '12 at 19:25
Thanks, I appreciate the answer. – Adelaide Dokras Mar 4 '12 at 10:00