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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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    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.2012-08-16

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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! +12012-08-16
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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 $\,x_n\to 0\,$