When do $p$-adic and standard convergence coincide?

Question

Consider the series $1 + 3/5 + 3^2/5^2 + \cdots = \sum_{j=0}^\infty 3^j/5^j$. In both the standard metric and the $3$-adic metric, $\sum_{j=0}^\infty 3^j/5^j = 5/2$. Is this a coincidence because of the geometric nature of the sum (which has an algebraic solution method that's valid in both fields), or is the case that if $x_n \to x \in \mathbb{R}$, and $x_n \to y \in \mathbb{Q}_p$, then $x = y$ (via the natural embedding)? If it's not the case in $\mathbb{Q}_p$, then what about in $\mathbb{Z}_p$?

Answer 1

First of all, there is no natural embedding from $\Bbb R$ to $\Bbb Q_p$ or vice-versa.

If you have a sequence of rational numbers, then it may converge both in $\Bbb R$ and in $\Bbb Q_p$, but it may converge to transcendental numbers in either field.

Example: $\sum_{n \geq 0} \frac{3^n}{n!}$ converges in $\Bbb R$ to $e^3$, and in $\Bbb Q_3$ to a transcendental number in $\Bbb Q_3$ (which is the analog of $e^3$). It is meaningless to say whether these two are equal.

In your example, it is special because the series you are using is the Taylor expansion of a rational function in $\Bbb Q$. It then makes sense, because both series converge to the value of the rational function, which takes values in $\Bbb Q$.

In general, suppose you have a sequence $(x_n)_n$ of rational numbers, which converges both in $\Bbb R$ and in $\Bbb Q_p$, and the limits are both rational numbers, they still may be different.

Example: $x_n = \frac {3^n}{1 + 3^n}$. In $\Bbb R$ it converges to $1$, while in $\Bbb Q_3$ it converges to $0$.