Convergence of series in p-adic field or non-archimedian field

Question

Field of $p$-adic numbers:

Show that if $ \ \sum_{n=0}^{\infty} a_n \ $ converges , then $ \ \sum_{n=0}^{\infty} a_n^2 \ $ converges in p-adic field.

Answer:

At first remember that this is a Non-Archimedian field.

So the comparison test may be different.

That is why I am getting no way to prove it because

$ \sum_{n=0}^{\infty} a_n \ $ converges means $ \ a_n \to 0 \ $

Thus $ \ a_n \geq a_n^2 \ $

By comparison test,

$ \sum_{0}^{\infty} a_n \geq \sum_{0}^{\infty} a_n^2 \ $ converges.

But here in Non-archimedian field or p-adic field the comparison test is applicable or not I don't know.

Can some one help me?

Answer 1

There is a perfectly valid $p$-adic comparison test, as uninteresting as it may be.

Namely, if $\sum a_n$ is convergent and for each $n$, $|b_n|\le|a_n|$, then $\sum b_n$ is also convergent.

But as @Watson has pointed out, if $\sum a_n$ is convergent, then $|a_n|\to0$, and thus under the hypotheses on $\{b_n\}$, you get $|b_n|\to0$, so that automatically $\sum b_n$ will be convergent.

Answer 2

In a non-archimedean space, we have the wonderful property that $\sum_{n \geq 0} a_n$ converges as soon as $a_n \to 0$. Since $a_n \to 0$ implies that $a_n^2 \to 0$, you can conclude that $\sum_{n \geq 0} a_n^2$ converges.