Why do we pick $\frac{1}{1-x}$ for $f(x)=1+x+x^2+x^3+...+x^n$, where $n$ tends to infinity?

Question

When we consider the function $f(x)=1+x+x^2+x^3+...+x^n$ where $n$ tends to infinity, we can rewrite this as $$f(x)=1+x(1+x+x^2+x^3+...)=1+x(f(x))\qquad (1)$$ After some algebraic manipulations, we arrive at $$f(x)=\frac{1}{1-x}\qquad (2)$$ This expression is used to assign a "sum" to certain divergent series. If we take another look at (1), however, we can also write $$f(x)=1+x(1+xf(x))$$ from which we can deduce that $f(x)=\frac{1+x}{1-x^2}$. Of course, we can produce infinitely many functions that satisfy $f(x)$ in this manner, but I only see (2) in the literature. Is there a good reason for this phenomenon? Thanks,

Max

Answer 1

$1-x^2=(1-x)(1+x)$ so you did not find a new solution.

Answer 2

Notice that

$$f(x) = \frac{1+x}{1-x^2} = \frac{1+x}{(1+x)(1-x)} = \frac{1}{1-x}$$

so you end up with the same function.

Answer 3

Let $f(x) = \sum x^k$ and let $g(x) = 1/(1-x)$. Then where $f$ makes sense (namely for $|x| < 1$) the functions $f$ and $g$ agree. This answers your last question "Is there a good reason for this phenomenon?"