Let $R$ be any ordered field. Define $R((x)):=\{\sum\limits_{k=-n}^\infty a_kx^k:a_k \in R \ , \ n \in\mathbb N\}$
Addition:
$\sum_\limits{k=-n}a_kx^k+\sum_\limits{k=-m}b_kx^k:=
\sum_\limits{k=-max\{n,m\}} (a_n+b_n)x^n$
Multiplication:
$\sum_\limits{k=-n}a_kx^k \times\sum_\limits{k=-m}b_kx^k:=
\sum_\limits{k=-(n+m)} \left(\sum_\limits{i+j=k}(a_ib_j)\right)x^k$
Order Relation: $\left(\sum_\limits{k=-n}^\infty a_kx^k\right) \geq 0$ iff the first nonzero coefficient is greater than or equal to $0$. Using this we can define $\alpha \leq \beta$ iff $0 \leq \beta-\alpha.$
Then $\left(R((x)),+,\times, \leq\right)$ is an ordered field.
Archimedean Property: By the order relation defined above, we can see that there exists $x^{-1} \in R((x))$ with $1<2<…<n<…<x^{-1}$ $\implies$ the field is non-archimedean.
Defining modulus function:
$\left|\sum_\limits{k=-n}a_kx^k\right|_{R((x))}:=\begin{cases}
\sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient postive} \\
-\sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient negetive}
\end{cases}$
Cauchy Sequences: $(\alpha_n)$ is said to be cauchy, if for every $(0<)\varepsilon \in R((x))$ there exists $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.
Now I want to see if $R((x))$ is Cauchy Complete, (Do I need to assume that $R$ is Cauchy Complete?)
Let $(\alpha_n)$ be a Cauchy Sequence in $R((x))$ that is for any $(0<)\varepsilon \in R((x))$ there exists an $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.
Taking the first nonzero coefficient from each $\alpha_n$ we can obtain a Cauchy sequence $(a_n)$ in $R$.
If I assume $R$ is Cauchy Complete then $(a_n) \to a(\in R)$.
But I don't know what to do next.
I am not sure whether the above definitions are meaningful. Any insight will be of great help.
Best Answer
Given $(a_n)$ Cauchy,
For all $K$ there is $N_K$ such that for $n\ge m\ge N_K$ either $0\le a_n-a_m < x^K$ or $0\le a_m-a_n < x^K$ which means that $a_n-a_m\in x^K R[[x]]$.
Writing $a_n = \sum_{j\ge J_n} A_{n,j} x^j$ we get that $(A_{n,j})_{n\ge 1}$ is constant for $n$ large enough and $$\lim_{n\to \infty} a_n = \sum_{j\ge \min(0,J_{N_1})} (\lim_{n\to \infty} A_{n,j}) x^j\in R((x))$$
(the topology is generated by the open sets $\{ f\in R((x)), b-\epsilon<f<b+\epsilon\}$)