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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 $\sum_{i=0}^{\infty} x_{i}$?

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What do you mean when you say "$x_i(j)$ is finite"? If I understand what you have written, $x_i(j)$ is a coefficient. Do you mean $\{i \mid \text{ only finitely many }x_i(j)\neq 0 \}$ ? –  rschwieb May 29 '12 at 16:21
    
@rschwieb: modified the text a little, hope it is clear now. –  user31509 May 29 '12 at 17:10
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Yes that helps a lot, thank you. It's also good to know we're assuming they are all finite. I was worrying about cases :) –  rschwieb May 29 '12 at 17:19
    
@rschwieb: please let me know if you have a proof. –  user31509 May 30 '12 at 0:19
    
I don't think I can improve on wxu's proof, except I could restate the main idea. Let's say you want to match $x$ on the first $n$ coefficients. All you would have to do is go up high enough that the first $n$ coefficients stop changing. You can do this by picking any $k$ above all indices involved in computing the first $n$ coefficients. When you achieve that, you are within epsilon of the limit. –  rschwieb May 30 '12 at 1:01

1 Answer 1

up vote 2 down vote accepted

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.

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