Can sum of a series be uncountable

This question confuses two notions of infinity. The notion of infinity in set theory, that describes how big a set is, is not the same as the infinite of analysis. In short, the infinity of analysis means simply "grows without bound." On the other hand, the infinity of set theory talks about what kind of set you can put your set in one-to-one correspondance with. A set has cardinality $\aleph_0$ when it can be put in such a pairing with the natural numbers.

The slightly longer story: A series is a limit. In particular, it is a limit of the sequence of partial sums of its terms. We say that the series converges when this sequence approaches some finite number. In particular, the sequence converges to the limit $S$ if for every positive real number $\epsilon$, there is an index of the sequence $N$, so that for every index larger than $N$, the terms of the sequence never get more than $\epsilon$ away from $S$. What this means is that the terms get and stay close to a particular number, and as we take $\epsilon \to 0$, we can get closer and closer and closer to this number.

When this doesn't happen, it can do one of a few things. It could bounce around and never settle down. The prototypical example is $(-1)^n$. Bounces back and forth, but never settles. Another possibility is that it blows up, with terms who's (absolute) values will eventually exceed any possible number. In this case, we say that the sequence 'diverges.' Since the infinite series is just a special kind of sequence limit, here too we say the series diverges when the partial sums are getting larger, and will grow larger than any finite number.

So while the infinity of analysis deals with a kind of growth and change, the infinity of set theory is rather stagnant, and measures a kind of size. Neither is better or worse, they just capture different notions.

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 $n = \aleph_0$.