Real Number Construction Would the sequence of partial sums  from indefinitely adding $1.0 \times 10^{-n}$ to zero, correspond to the set of positive real numbers?
 A: No, not all real numbers can be expressed as $\sum_{i=1}^k10^{-n}$. This is only the set of real numbers with a finite decimal expansion. For instance, the number $\frac{1}3=.333333\ldots$ is not a finite sum of terms of the form $10^{-n}$ since we have the bounds $$\frac{10^n-1}{3}\cdot 10^{-n}<\frac{1}3<\frac{10^n+2}{3}\cdot 10^{-n}$$
where the terms $\frac{10^n-1}3$ and $\frac{10^n+2}3$ can be seen to be consecutive integers, meaning $\frac{1}3$ is always between two of the partial sums. Another way to write this inequality is:
$$0.\underbrace{33\ldots33}_{n-2\text{ times}}3<\frac{1}3<0.\underbrace{33\ldots33}_{n-2\text{ times}}4.$$
That said, numbers with finite decimal expansions are "dense" in the reals, which means that they can approximate any other number arbitrarily well. That is, if we chose some real number $x$ and some value $\varepsilon>0$ we can find a suitable number with finite decimal expansion which differs from $x$ by no more than $\varepsilon$. For instance, if we want a number that approximates $x=\pi$ to within $\varepsilon=10^{-4}$, we could see that the number $3.1415$ satisfies that and can be written of the desired form.
A: The vital missing pieces of information in the original question statement
are the assumptions implied by this comment:

Hasn't every number got infinite digits, 1 = 1.0... = 1.0' and wouldn't the sequence of partial sums be uncountably infinite when n is infinity? R0 = {0.0 x 10^-n}, R1 = {0.0 x 10^-n + 1.0 x 10^-n}, ...

Yes, every number can be said to have infinitely many digits.
The question is, what is $10^{-n}$ if $n$ is infinite?
Whatever it is, it is not the smallest possible positive real number,
because there is no such thing.
A reasonable interpretation would be that by "$10^{-n}$ where $n$ is infinite" you actually mean the limit of $10^{-n}$ as $n$ goes to infinity.
This is evaluated as follows:
$$ \lim_{n\to\infty} 10^{-n} = 0.$$
This does not produce the sequence you want, because you can add $0$ to
$0$ forever and it will always remain $0$.
In any case, there is no $n$ such that you can get "the next" real number by adding $10^{-n}$ to any real number, because there is never a unique "next" real number. Suppose $R_0$ is a real number and $R_1$ is "the next" real number greater than $R_0$. Then $x = (R_0 + R_1)/2$ is also a real number, but $R_0 < x < R_1$, so $R_1$ is not "the next" real number greater than $R_0$. 
The best you can do by constructing any increasing sequence of partial sums is to generate a countable subset of the real numbers with uncountably many "missing" real numbers between each pair of numbers in your sequence.
