Metric on ring of formal power series Consider the ring of formal power series $R[[x]]$ and given $\sum a_{n}x^{n}$ define a metric on $R[[x]]$ as follows: $d((a_n),(b_n))=2^{-k}$ where $k$ is the smallest natural number such that $a_{k} \neq b_{k}$ (if no such $k$ exists we define their distance to be zero).
Let $\{x_{i}: i \in \mathbb{N}\}$ be a sequence of elements of $R[[x]]$ (i.e each $x_{i}$ is a formal power series). Now for each natural number $j$ and each sequence $\{x_{i}: i \in \mathbb{N}\}$ let $G(x_{i},j)$ be the set of all natural numbers i such that $x_{i}(j) \neq 0$.
For example $x_{1}$ is a formal series, and $x_{1}(j)$ means the $j$-th coefficient of the formal series $x_{1}$. 
From now on assume $G(x_{i},j)$ is finite.
Define a formal series $\sum_{i=0}^{\infty} x_{i}$ as follows:
$(\sum_{i=0}^{\infty} x_{i})(j):= \sum_{i \in G(x,j)} x_{i}(j)$, i.e the $j$-th coefficient of $\sum_{i=0}^{\infty} x_{i}$ is equal to $\sum_{i \in G(x,j)} x_{i}(j)$.
Question: let $\{x_{i}: i \in \mathbb{N}\}$ be a sequence of formal power series and let $S_{n}=\sum_{i=0}^{n} x_{i}$. Is it true that the sequence $\{S_{n}: n \in \mathbb{N}\}$ converges as $n \rightarrow \infty$ (with respect the metric defined previously) to $\sum_{i=0}^{\infty} x_{i}$?
 A: This is my guess.
"Now suppose for each natural number $j$, the set $\{i \mid x_i(j)\neq 0\}$ is finite, which we denote by $G((x_i),j)$, where $(x_i)$ denote the sequence $x_1,x_2,\ldots$" OK, if it is in this case, the answer is yes.
Let $y$ be the formal series $\sum_{i=0}^{\infty} x_{i}$ defined by the context. Then for any $k\geq 0$, we may find an $N$, such that for $n\geq N$, $|S_n-y|\leq 2^{-k}$. For example, our $$N=Max\{i\mid i\in \bigcup_{j=1}^k G((x_i),j)\}+2.$$

Edit: What does the number $M=Max\{i\mid i\in \bigcup_{j=1}^k G((x_i),j)\}$ mean? For example if it is one. Then $x_l(j)=0,j=1,2,\ldots, k$ for $l\geq 2$, hence $S_1(1)=x_1(1)=x(1),S_2(1)=x_1(1)=x(1),\ldots,S_l(1)=x(1),$ for any $l\geq 1$. And so on. 
Maybe a more concrete example say everythings. Let $x_1=1,x_2=x,\ldots,x_n=x^{n-1},\ldots$. Then $G((x_i),0)=\{1\},G((x_i),1)=\{2\},G((x_i),j)=\{j+1\}$. The number $Max\{i\mid i\in \bigcup_{j=1}^k G((x_i),j)\}=Max\{i\mid i\in\{0,1,2,\ldots,k+1\}\}=k+1$. And $S_n=1+x+\ldots+x^{n-1}$, $|S_n-y|=2^{-n}\leq 2^{-k}$ if $n\geq k+1\geq k$.
Sorry, the index makes me tired.
