Closed form of generating function consisting of power of two binomials

Question

Let $g(x)$ be infinite formal power series and $$g(x) = (1 + x)(1 + x^2)\cdots(1 + x^{2^k})\cdots$$ Show that $(1 - x) g(x) = 1$. My book gives following proof:

Using a fact that $(1 - x^k)(1 + x^k)=1-x^{2^k}$

$$ (1 - x) g(x) = (1 - x)(1 + x)(1 + x^2)\cdots(1 + x^{2^k})\cdots = (1 - x^2)(1 + x^2)\cdots(1 + x^{2^k})\cdots =$$ $$(1 - x^4)(1 + x^4)(1 + x^8)\cdots = (1 - x^8)(1 + x^8)(1 + x^{16})\cdots = 1 $$

Each step of reasoning shows that every term is "eaten" by multiplication. I do not see how it equates to 1 and I am not even close to accept it as a formal enough proof. If it was regular power series (numeric not formal) then I see how this proof could be easily turned in formal one by using limit but I do not see why formal power series of this form equates to $1$.

I have added logic tag because I do not understand proof technique used here. It is not any "inifitiary" proof technique which is widely used like limits or induction.

Answer 1

What we really want to understand is why these sorts of formal manipulations are valid. To do so it will suffice to convince ourselves that for all of the coefficients of $x^i$ in $(1-x)g(x)$ where $i<n$ for any fixed $n$ we have a coefficient of $0$. By using the "eating" process you can show that the coefficient of each $x^i$ with $i>0$ in this product is $0$. By inductively repeating the process, you discover all of the coefficients of the formal power series, so you know what the formal power series is. Then, note that every time we "ate" the previous term, the constant term, $1$, is preserved, so, inductively, we see that the constant term will always be $1$, and thus that $(1-x)g(x)=1$.

Answer 2

In the formal power series ring $R[[X]]$, what does it mean for a sequence of elements to converge?

This is the question you should be asking yourself before you even tried to look at the proof. The answer is that it means for every $i\ge0$, the sequence of $X^i$ coefficients of $f_1(X),f_2(X),f_3(X),\cdots$ will eventually be constant, say $a_i$. The series it converges to is then $\sum_{i\ge0}a_iX^i$.

The partial product is $(1-X)(1+X)(1+X^2)\cdots(1+X^{2^k})=1-X^{2^{k+1}}$. Clearly then, if $i>0$ the coefficient of $X^i$ in the partial products will eventually be constantly $0$ (for $i>2^{k+1}$) and if $i=0$ the term is constantly $1$. Hence the infinite product is $1+0X+0X^2+\cdots=1$.

Induction is implicit in the simplification of the partial products. Limits are implicit in how we define convergence of sequences of formal power series. There is even a topology lurking underneath all of this called the $(X)$-adic topology. Here, a neighborhood basis of the identity $0\in R[[X]]$ is given by the nested decreasing sequence of ideals $(X)\supset(X^2)\supset(X^3)\supset\cdots$. This construction is a special case of more general ones, that of completions and more generally inverse limits of rings and groups. It is also called the Krull topology (for example in the definition of an absolute Galois group of a field). And a different generalization is through valuation rings.