Is that a convention that $\frac{1}{1-x}=1+x+x^2+...$ in $\mathbb{Q}_p$ or $p$-adic number theory?

Question

I saw $\frac{1}{1-x}=1+x+x^2+...$ involving $\mathbb{Q}_p$ on some notes, but in calculus, it works only $|x|<1$, but in these notes, $x$ is not supposed to have this restriction. I wonder if it's a convention, in other words, if it's defined that the inverse of $(1-x)$ is defined to be $1+x+x^2+...$?

Answer 1

In $\mathbb{Q}_p$, a series $\sum\limits_{i\geq 0}a_i$ converges if and only if $|a_i| \rightarrow 0$ (this only hold in one direction in real analysis, for example if $a_i=\frac{1}{1+i}$ then this doesn't converge in real analysis).

So $\sum\limits_{i \geq 0} x^i$ converges if and only if $|x^i| \rightarrow 0$, if and only if $|x|<1$, which is the same criterion as for real analysis, except this time the norm is taken $p$-adically (loosely speaking, the requirement is that $p|x$).

The equality does not hold for other elements, take for example $x=1$, where the sum should be $\infty \not\in \mathbb{Q}_p$, or even $x=-1$ where the sum makes absolutely no sense at all (either in $\mathbb{R}$ or $\mathbb{Q}_p$).

Answer 2

You slightly misunderstand the nature of the statement about $\sum_{i=0}^\infty x^i$. The theorem you're used to says

If $\sum_{i=0}^\infty x^i$ converges, then it converges to $\frac{1}{1-x}$

It happens to be that, in $\mathbb{C}$, this occurs when $|x|<1$, but the theorem is provably solely off the assumption of convergence. Likewise, in $\mathbb{Q}_p$, if $\sum_{i=0}^\infty x^i$ converges, then it converges to $\frac{1}{1-x}$. In $\mathbb{Q}_p$ the criterion for that sum converging is no longer $|x|<1$. A sequence $\alpha_i\in\mathbb{Q}_p$ converges iff $|\alpha|_p\to 0$, so in $\mathbb{Q}_p$ we have:

$$\lim_{i\to\infty}|x^i|_p=0\iff \sum_{i=0}^\infty x^i=\frac{1}{1-x}$$