Prove R[[x]] is a complete metric space. I am attempting to show that $R[[x]]$ (the ring of formal power series) is a complete metric space, where $d(A,B)= 2^{-ord(A-B)}$ if $A\neq B$ and $d(A,B) = 0$ if $A=B$.
I have shown that it is a metric space but am struggling with proving it is complete. I know that I need to show that every Cauchy sequence converges to something in $R[[x]]$ but am not sure how to go about this.
Thank you for any help!
 A: Let $N$ be a non-negative integer. If $d(A,B) < 2^{-N}$ then $ord(A-B) >N$ which implies that the first N coefficients in A and B are equal. If $\{A_k \}$ is a Cauchy sequence then $d(A_n,A_m) < 2^{-N}$ for all n and m sufficiently large so the first N coefficients in $A_n$ and $A_m$ are the same for all n and m sufficiently large. In particular the N-th coefficients in $A_1,A_2,...$ become the same after some stage.
Denoting by $K_i(A_n)$ the $i$-th coefficient in $A_n$, then one can define
$$a_i=\lim_{n\to +\infty}{K_i(A_n)}.$$
This way you obtain $a_0,a_1,a_2,...$. Now let A be the formal power series with these coefficients. Then $d(A_n,A) \to 0$. To see this let $\epsilon >0$ and choose N such that $2^{-N} <\epsilon$. It is cleat from the definition of $a_n$'s that for all n sufficiently large the first N coefficients of $A_n$ and A coincide. This implies $ord(A_n -A) \geq N$ and $d(A_n,A) <2^{-N} <\epsilon$ for all n sufficiently large. We have proved that $A_n \to A$.
A: Just adopt one of the notations which consists in "fetching the coefficient of $x^n$ in a series $S$ ". Then, formally, for $S=\sum_{n\geq 0}a_n\,x^n$, one denotes $[x^n]S:=a_n$. Now, let $S_n$ be a Cauchy sequence of $\mathbb{R}[[x]]$, it means that 
$$
(\forall \epsilon>0)(\exists N)(p,q\geq N\Longrightarrow d(S_p,S_q)<\epsilon) 
$$
from this, you see that, for all fixed $k\in \mathbb{N}$, the coefficient $[x^k]S_n$ is stationary after a certain value of $n$. Then define $l_k=lim_{n\to\infty}[x^k]S_n$ and you see that $S_n$ converges to 
$$
L:=\sum_{k\geq 0}l_k\,x^k
$$
Hope this is of help.
