TITLE: Can sum of a series be uncountable
QUESTION [12 upvotes]: There are several methods to say whether sum of series is finite or not. Can we say whether sum of series is countable or not.
For example $S_n=\Sigma_{0 \leq i \leq n}{2^i}= 2^{n+1}-1$ So for $n=\aleph_0$
$ S_{\aleph_0}=2^{\aleph_0}-1$ 
So can we say that S has a value which is not countable. But wouldn't that mean that sum of integers turns out be a non integer, as all integers form a set of countables. Or can we say that the series $S_i$ has a limit which lies outside set of inetgers so $\{S_1, S_2, S_3 ...\} \subset I$ has a limit $p$, such that $p \notin I$
Edit: I understood that $\{S_1, S_2, S_3 ...\} \subset I$ has a limit $p$ is incorrect as neither $\aleph_0$ nor $2^{\aleph_0}$ lie in $\mathbf{R}$ but I am still confused about other things. 
Edit: Suppose we have sets $A_1, A_2, ... A_i...$ each having $2^i$ elements  for $i=1,2 ...$ i.e. $\#(A_i)=2^i$. Suppose all these sets are disjoint can we say whether the set $A=\bigcup_{1 \leq  i \leq \infty} A_i$ is countable or not.

REPLY [14 votes]: Recall that the statement $S_n = 2^{n+1} - 1$ is proven by induction. Induction is not magic - it cannot apply to things that aren't in its domain. For example, just because $S_n = 2^{n+1} - 1$ doesn't mean $S_{\mathrm{apple}} = 2^{\mathrm{apple}+1}-1$. Induction operates in two steps: first, the base case shows that the claim holds at $n = 0$. Second, the induction step shows that if the claim holds for $n$ then it holds for $n+1$. Now, to show that (for example) it works for $10$, you notice that the induction step says that if it works for $9$ then it works for $10$. If it works for $8$, then it works for $9$. And so on and so forth, back to $0$, which we already know works.
The thing is that for apples or $\aleph_0$, there isn't a way to step back to $0$ one by one. The argument by induction doesn't work, and so the final result has no reason to work either.
EDIT: You mentioned in a comment the equation $(x - 1)(x^n + x^{n - 1} + \ldots + 1) = x^{n+1} - 1$. This equation is proven by induction on $n$. It therefore holds for all natural numbers, but $\aleph_0$ is not covered by the induction. I'll draw your attention, for example, to the case where $n$ is negative, or $n = \pi$. The equation doesn't even make sense for those values of $n$, because those values are too different from positive natural numbers. The same is true of $\aleph_0$.
As I mentioned in a comment below, virtually nothing that is true of finite $n$ is also true of $\aleph_0$. Even the simple statement $n < n + 1$ is false when $n = \aleph_0$.