Let $R$ be a semi-ring (resp. a ring).
Let $\hat{R}$ be the structure obtained by "allowing infinite sums" in $R$ so 
$$\hat{R} := \lbrace\sum\_{i \in I} a\_i| a\_i \in R\rbrace,$$
where $I$ is countable set.
 So my question is:


1. Is $\hat{R}$ a semiring (resp. ring)?
2. If we assume $R \neq 0$ does it follow that $\hat{R} \neq 0$?


Edit: After being closed, I rewrite the question as follows: 
 Given a semiring $R$. Assume we would like to make sense of infinite sums in $R$, so is there a way to define a sort of "completion" to $R$ which allow to make sense of infinite sums?
Thanks. 



On rings and semirings
It was pointed out in the comments that if we mean by "$\sum\_i a\_i$" just the $I$-tuple $(a\_i: i \in I)$, then the family of such tuples can of course be made into a semi ring (e.g., by pointwise addition and multiplication, or Cauchy multiplication if you use the index set $\mathbb N$). But in this case, neither the notation/name "sum" makes sense to me, nor does question 2 make sense: $R^I$
is never empty for nonempty $R$. 


So I will reinterpret the question: Which semirings can be extended to complete semirings? By "complete semiring" I mean a semiring $(S,+,\cdot,0,1)$ (with commutative addition) in which all sums $\sum\_{i\in I} a\_i$ (for all index sets $I$, or at least for all countable $I$) are defined and satisfy reasonable laws, in particular: 


* infinite associativity and commutativity
* left distributivity: $c\cdot \sum\_i a\_i = \sum\_i c\cdot a\_i$, and similarly right distributivity; in particular, the sum of 0s is 0.
* finite sums agree with addition.


If you also impose the following (natural, I think) condition, called "d-completeness" or "discrete completeness": 


* Whenever $\sum\_{i=0}^n a\_i = a\_0$ for all $n\ge 0$, then also 
$\sum\_{n=0}^\infty a\_i = a\_0$


then it is easy to see that the following are equivalent. 


1. The semiring $S$ is a subsemiring of a complete semiring.
2. $S$ can be partially ordered by $a \le b$ iff $\exists x\ a+x=b$.
3. For all $a,b,c$ in $S$, $a+b+c=a$ implies $a+b=a$. (Of course, as Darij Grinberg has pointed out, this is never true in rings.)


The point is that $a+(b+c)+(b+c)+\cdots = a+b + (c+b)+(c+b)+\cdots$. A few details can be found [here](http://arxiv.org/abs/math/0208134).  

Georg Karner's paper "On limits in complete semirings" (Semigroup Forum 45 (1992), no. 2, 148–165, MR1171841 (93h:16078)) has more information, and uses topological concepts rather than order-theoretic ones. 


