Why isn't $1 + x + x^2 \dots = \frac{1}{1-x}$?

Question

In example 8.1 of "Combinatorics and Discrete Mathematics", we consider the generating function: $$ F(x) = \sum_{n=0}^\infty x^n.$$ We find that $$(1 + x + x^2 + \dots) - x(1 + x + x^2 + \dots) = 1,$$ which implies that $$(1-x)F(x) = 1.$$ Finally getting $$F(x) = \frac{1}{1-x}.$$ Now, the series only converges when $x < 1$. But in the rewriting steps that we just used, we didn't actually use any calculus, or any other techniques which make this restriction clear. So I would think that $F(x) = \frac{1}{1-x}$ as long as you don't divide by zero.

But this is obviously not the case, since the sum clearly diverges. What am I misunderstanding here?

Answer 1

But in the rewriting steps that we just used, we didn't actually use any calculus

On the contrary, you did use calculus, you just hid it in the symbol "$\dots$". What does that symbol really mean? Well, in general, when we write

$$a_1 + a_2 + a_3 + \cdots$$ what we really mean is $$\lim_{N\to\infty}(a_1 + a_2 + \cdots + a_N)$$

and there is your calculus! Limits!

When operating with limits, you need to be very careful. In particular, the equation $$(1 + x + x^2 + \dots) - x(1 + x + x^2 + \dots) = 1$$ is a statement about limits, and is only valid if $|x|<1$.

Answer 2

On the one hand, when $x$ is a real number, $1 + x + x^2 + \cdots$ only makes sense when $|x|<1$.

On the other hand, in a ring of formal power series $R[[x]]$, the polynomial $1-x$ has an inverse and it's $1 + x + x^2 + \cdots$. No convergence enters here.