I need a proof for this:
Let $(F_j)_{j\ge 0}$ be a sequence of formal power series.
The infinite product $\prod_{j\geq0}(1+F_j(x))$, where $F_j(0)=0$, converges if and only if $\lim_{j\to\infty}\deg F_j(x)=\infty$.
In this case, the series $G_I(x)=\prod_{i\in I}F_i$, with $|I|< \infty $ is summable and $\prod_{j\geq0}(1+F_j(x))= \lim_{n \to \infty} \prod_{j=0}^n (1+F_j(x)) = \sum_{|I| \leq 0}G_I(x)$.
I don't even know where to begin. Thanks in advance.
It’s just a matter of thinking of things systematically. You have your infinite product, let’s think of it as $(1+F_1)(1+F_2)(1+F_3)\cdots\,$, but let’s write the factors beneath each other, one per line. Now: what does the infinite product look like if it’s expanded? You certainly have one factor that’s gotten by running down the left-hand column, all $1$’s. This is your $\sum_{|I|=0}G_I$. Then look at how many ways there are of going all the way down that left column, but taking just one “monomial” $F_i$ from the second column. Of course the totality of these is $F_1+F_2+F_3+\cdots\,$. This is your $\sum_{|I|=1}G_I$. Now how many ways can you get a contribution where you take two factors from the second column? You’ll have all possible $F_iF_j$’s, added up, to get something like $F_1F_2+F_1F_3+F_1F_4+F_2F_3+\cdots\,$. This will be your $\sum_{|I|=2}G_I$. And so it goes. All you need to do is fill in the details. It may help to think about what the full expansion is when there are only finitely many of the $F$’s.
