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EDIT: Let R be a commutative ring with unit ring and $I$ a maximal ideal in R. The completion of R with respect to $I$ is the inverse limit of the factor rings $R / I^k$ under the usual quotient maps.

A ring is said to be complete with respect to a maximal ideal if the map to its completion with respect to that ideal is an isomorphism.

See this page for additional info on convergence and ring completions.

Let $R$ be a ring such that it is complete w.r.t. some ideal $I$ and let $(x_n)_n$ be a sequence in $I$. I was told that $$\sum_{n=1}^\infty x_n\quad\mbox{converges}\iff x_n\mbox{ converges to }0.$$ The direction $\implies$ is trivial, the other direction is harder. How can I prove this part?

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Would you care to give the definition/some link as to what "ring complete wrt some ideal" means so that infinite series and limits of sequences in it make sense? It seems to be that somehow the above introduces some topology on the ring...? –  DonAntonio Aug 16 '12 at 16:12
    
@DonAntonio I belive he's referring to the completion of a ring with the I-adic topology. –  rschwieb Aug 16 '12 at 16:26
    
I think you may be right, @rschwieb. Thanks –  DonAntonio Aug 16 '12 at 16:33
    
In the case of $R = \mathbb{Z}$, $I = (p)$ for a prime number $p$, you can use the ultrametric property of the $p$-adic norm. –  Zhen Lin Aug 16 '12 at 16:37

2 Answers 2

Working with inverse limits makes me feel relatively sure (had to deal quite a bit with profinite groups some years ago). Now, what you want to prove is true for the p-adic topology, which makes analysis of series there much simpler than with the usual completion of the rationals to the reals, and copying the arguments there works here, too:

Here, we can say $\,\{x_n\}\,$ is a Cauchy seq. iff $$\lim_{n\to\infty} (x_{n+1}-x_n)=0$$ and since we have the ring is complete then the series converges iff its partial sums' sequence is a Cauchy sequence iff it fulfills the above condition, which of course is equivalent to $\,x_n\to 0\,$

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up vote 4 down vote accepted

I've just figured it out and for those interested, here is my proof:

Note that in any complete ring, a series converges if and only if the sequence of its partial sums $(s_n=\sum_{k=1}^nx_k)_n$ converges.

Since $(x_k)$ converges to $0$, there is a $M\in\mathbf N$ such that $x_n\in I^k$ for each $n\ge M$. So by choosing $N = M$,$$s_j-s_n=\sum_{i=1}^jx_{j+i}\in I^k$$ for each $j,n\ge N$.

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Oh, this is just what I wrote...nice! +1 –  DonAntonio Aug 16 '12 at 16:43

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