Show that the series converges

[evinda]

Hello! (Wave)

How can I show that in the field $\mathbb{Q}_p$ stands the following? (Thinking)

If $\{ a_n \}_{n \in \mathbb{N}}$ a sequence of elements in $\mathbb{Q}_p$ and $\lim_{n \to \infty} a_n=0$, then the series $\sum_{n=1}^{\infty} a_n$ converges.

 

[Euge]

Hello! (Wave)

How can I show that in the field $\mathbb{Q}_p$ stands the following? (Thinking)

If $\{ a_n \}_{n \in \mathbb{N}}$ a sequence of elements in $\mathbb{Q}_p$ and $\lim_{n \to \infty} a_n=0$, then the series $\sum_{n=1}^{\infty} a_n$ converges.

Let $\varepsilon > 0$. Choose $n_0$ such that $|a_n|_p < \varepsilon$ for all $n \ge n_0$. Then for all $N$ and $k$ with $N > k \ge n_0$,

\(\displaystyle \left|\sum_{n = k}^N a_n\right|_p \le \max\{|a_k|_p,|a_{k+1}|_p,\ldots, |a_N|_p\} < \varepsilon\)

What can you deduce from this?

 

[evinda]

Let $\varepsilon > 0$. Choose $n_0$ such that $|a_n|_p < \varepsilon$ for all $n \ge n_0$. Then for all $N$ and $k$ with $N > k \ge n_0$,

\(\displaystyle \left|\sum_{n = k}^N a_n\right|_p \le \max\{|a_k|_p,|a_{k+1}|_p,\ldots, |a_N|_p\} < \varepsilon\)

What can you deduce from this?

Could you explain me why we take the sum from $n=k$ till $n=N$ ? (Thinking)

 

[Euge]

Could you explain me why we take the sum from $n=k$ till $n=N$ ? (Thinking)

The point is to show that the sequence of partial sums of the series \(\displaystyle \sum_{n = 1}^\infty a_n\) is Cauchy in \(\displaystyle \Bbb Q_p\).

 

[evinda]

The point is to show that the sequence of partial sums of the series \(\displaystyle \sum_{n = 1}^\infty a_n\) is Cauchy in \(\displaystyle \Bbb Q_p\).

So is it like that? (Thinking)
From $\lim_{n \to +\infty} a_n=0$ we get that , $\forall \epsilon>0 \ \exists n_0$ such that $\forall n \geq n_0$: $|a_n|_p< \epsilon$. Then , $\forall N>k \geq n_0$, we have:

$$\left |\sum_{n=1}^N a_n-\sum_{n=1}^k a_n \right |_p=\left | \sum_{n=k}^N a_n\right |_p \leq \max \{ |a_k|_p, |a_{k+1}|_p, \dots, |a_N|_p \}< \epsilon$$

Then , knowing that the sequence of partial sums of the infinite series converges in $\mathbb{Q}_p$,
from the Cauchy Criterion, we have that the infinite series converges in $\mathbb{Q}_p$.

(Thinking)

 

[Euge]

So is it like that? (Thinking)
From $\lim_{n \to +\infty} a_n=0$ we get that , $\forall \epsilon>0 \ \exists n_0$ such that $\forall n \geq n_0$: $|a_n|_p< \epsilon$. Then , $\forall N>k \geq n_0$, we have:

$$\left |\sum_{n=1}^N a_n-\sum_{n=1}^k a_n \right |_p=\left | \sum_{n=k}^N a_n\right |_p \leq \max \{ |a_k|_p, |a_{k+1}|_p, \dots, |a_N|_p \}< \epsilon$$

Then , knowing that the sequence of partial sums of the infinite series converges in $\mathbb{Q}_p$,
from the Cauchy Criterion, we have that the infinite series converges in $\mathbb{Q}_p$.

(Thinking)

That's good, but there are two small errors. The expression $|\sum_{n = k}^N a_n|_p$ should be $|\sum_{n = k+1}^N a_n|_p$, and so $|a_k|_p$ should be omitted in $\max\{|a_k|_p, |a_{k+1}|_p,\ldots, |a_N|_p\}$.

 

[evinda]

That's good, but there are two small errors. The expression $|\sum_{n = k}^N a_n|_p$ should be $|\sum_{n = k+1}^N a_n|_p$, and so $|a_k|_p$ should be omitted in $\max\{|a_k|_p, |a_{k+1}|_p,\ldots, |a_N|_p\}$.

Nice, thank you very much! (Smirk)

 

[mathbalarka]

I should add here that this property can readily be generalized using a little topology. Not only this sequence you mention converge in $\mathbf{Q}_p$, but in fact all Cauchy sequence does, i.e., $\mathbf{Q}_p$ is a *complete* metric space.

To see how actually this happens, you need to see that $\mathbf{Q}_p$ can be realized as a metric completion of $\mathbf{Q}$ under the norm $|\bullet|_p$, instead of seeing it just as $\text{frac} \, \mathbf{Z}_p$, $\mathbf{Z_p}$ being $\lim \limits_{\longleftarrow} \mathbb{Z}/p^i$. The algebraic and topological definitions are indeed "isomorphic" in some sense but that requires some work (you can try it if you want)

Completion of a metric space $(X, d)$ in general is defined by adjoining to $X$ the limits of it's Cauchy sequence, denoted as $\overline{X}$. This is almost nearly (but not quite) obvious to see that $\overline{X}$ is complete, if you know a bit of point-set topology.

Note : This is merely a comment, not an answer to the original question asked.