I'm learning about Berkovich Spaces from scratch, and the building blocks (algebro-analytic) are the (generalized) Tate algebras in one variable (The material where I'm learning covers first the one variable case, then they use $n$ variables). I'm stuck in proving that this algebra is complete with respect to the Gauss norm. Here is the background needed:
Let $(k,|-|)$ be a complete non-archimedean field. Pick $r>0$. The Tate algebra is defined as:
$$k\{r^{-1}T\}:=\left\{f=\sum_{i=0}^{\infty}a_{i}T^{i}\in k[\![T]\!]: \lim_{i\rightarrow\infty}|a_{i}|r^{i}=0\right\},$$ so the elements are power series which converge on the closed disk of radius $r$. For an element $f=\sum a_{i}T^{i}\in k\{r^{-1}T\}$ we put $|f|_{\text{gauss}}:=\max_{i}|a_{i}|r^{i}$. The function $|-|_{\text{gauss}}$ is a multiplicative norm on $k\{r^{-1}T\}$. The main result is that $(k\{r^{-1}T\},|-|_{\text{gauss}}\})$ is complete. I have already proved this when $r\geq 1$, since in that case, for any Cauchy sequence $(f_{n})=(a_{n,0}+a_{n,1}T+a_{n,2}T^{2}+\cdots)$ in $k\{r^{-1}T\}$, we have the following bound:
\begin{equation}\label{eq1}|a_{n+1,i}-a_{n,i}|\leq |a_{n+1,i}-a_{n,i}|r^{i}\leq |f_{n+1}-f_{n}|_{\text{gauss}} \qquad (I)\end{equation} for each fixed $i$ and every $n$, thus for each $i$, the sequence $(a_{n,i})$ is a Cauchy sequence in $k$, and since the field is complete, there is a unique element $a_{i}\in k$ such that $\lim a_{n,i}=a_{i}$ for each $i$, then defining $f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots$ one can prove that $f\in k\{r^{-1}T\}$ and $\lim f_{n}=f$ with respect to the Gauss norm.
The same strategy doesn't work if $r<1$, since in that case we have $r^{n+1}<r^{n}$ for every $n\in\mathbb{N}$, therefore we don't have a bound like (I) so we can't proceed as the case $r\geq 1$.
I have done several attempts, but I couldn't prove completeness in this case.
Any hint would be apreciatted, Thanks