Can a number have an uncountably infinite amount of digits? I'm an extreme mathematical layman, so please excuse the probable ignorance and awkward phrasing of this question.
Is there such thing as a kind of number which has an uncountably infinite amount of digits after the decimal point?
To give an example of what I mean, say such numbers exist and we'll call one of them S. The digits of S are such that every rnth place (and rn can be any positive real number) corresponds to the first decimal place of sin(rn).
The 0th place of of S is 0. What follows are infinitely many zeros until the digit place that corresponds to sin^-1(.1) is reached, and the digit of that place will obviously be 1. Long after that, the .5th place is 4, the .61999th place is 5, the 1st place is 8, the πth place is 0, the 6000π/e-th place is 7.
So, S will have a rather dull repeating pattern of 0.000...111...222...333..., eventually returning to 0s at the πth place and then cycling on again. 
Another such number could be C, which has a similar rule but the rnth place corresponds to the first decimal of cos(rn) instead. C would look like 1.999...888...777, and so on.
I know S and C can't be real numbers (if these can even be considered numbers at all), because the cardinality of a real number's digits is equal to the set of the natural numbers. The cardinality of the digits of S and C is equal to the set of real numbers.
Now, is all this just useless mumbo-jumbo hiding behind ellipses?  Supposing it's not, what would the cardinality of the set of all these weird numbers like S and C be? Would it be larger than the set of all real numbers?
 A: Decimal representations of a number is just a representation - ie, it is a way of specifying a number by giving a certain amount of information about it.
In the case of real or complex numbers, since there are strictly more than $\aleph_0$ many of them, you need $\aleph_0$ (countably-infinite)-many digits to describe an arbitrary real/complex number. Similarly, for rational numbers, since there are $\aleph_0$ many of them, you only need finitely many digits to specify any rational number (note that non-terminating rational numbers must repeat their digits, so you only need to know the digits up to the first repeat).
There do exist mathematical structures (fields, rings, vector spaces, groups) that have cardinality greater than $|\mathbb{R}|$, in which case you will need at least $|\mathbb{R}|$-many digits to specify an element of that structure.
See for example:
https://mathoverflow.net/questions/44705/cardinalities-larger-than-the-continuum-in-areas-besides-set-theory
A: A number can be seen as an infinite sequence of digits (or bits, if you write in binary). In other words, it is an infinite word on the alphabet $\{0, \dotsm, 9\}$ (or $\{0, 1\}$ if you write in binary).
Now, a theory of words over ordinals and even words over linear orders has been proposed in [1].
Given a finite alphabet $A$ and a totally ordered set $I$, a word $(a_i)_{i \in I}$ is simply a function from $I$ to $A$. Usual finite words are the words indexed by finite orderings $I = \{1,2,...,n\}$. An infinite word is a word over $I = \mathbb{N}$, but you can define words over $\mathbb{R}$ if you wish.
This definition and the main results about these words belong to automata theory, which is probably not what you are looking for, but  you can have an "uncountably infinite amount of digits" in this way. 
[1] A. Bès and O. Carton, A Kleene theorem for languages of words indexed by linear orderings, Int. J. Found. Comput. Sci., vol. 17, no. 3, pp. 519-542, 2006.
A: If you want each element of your field $F$ to be represented by a unique sequence $\alpha \rightarrow \mathbb{N}$ (where $\alpha$ is the number of digits you require), its cardinal must be $2^{\alpha}$.
It seems to me that the field of surreal numbers with birth date $< \omega_1$ could satisfy your condition. Its cardinal is $(2^{\aleph_0})^{\aleph_1} = 2^{\aleph_1}$. I don't know if it's stricly greater than $2^{\aleph_0}$.
A: An uncountable sum cannot converge in a standard sense of the word. Let $\{a_j: j \in J\}$ where $J$ I an uncountable index: Define $K_n:=\{ a_k \in a_j : a_k >1/n\}$. Then, by a cardinality argument, one of the sets $K_n$ will have infinitely -many terms, and the infinite sum $\sum K_n > 1/n+ 1/n +\cdots$ will diverge. So there must be a countable index beyond which all terms must be zero, unless you have a different notion of convergence. 
