$$S_n$$ gets ugly to compute the hard way -- imagine the pain if n = 1000. And yet most of all counting numbers are actually larger than 1000. So I want to prove that:
$$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ for all counting numbers, n.
First let me prove that $$S_1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$. The proof writes itself: $$S_1 = \frac{9}{10} = \frac{9}{10} \times 1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$.
Next, I want to prove that if for a particular counting number k, that $$S_k = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}}$$ is true, that $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ follows as a result. So from the definition of the family, I know that $$S_{k+1} = S_k + \frac{9}{10^{k+1}}$$ and from assumption: $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \frac{9}{10^{k+1}}$$, So we have $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \; \frac{9}{10} \times \frac{1}{10^k} = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + 10^{-k} \right) = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + \frac{10^{-k} - 10^{-k-1}}{1 - 10^{-1}} \right) = \frac{9}{10} \times \frac{1 - 10^{-k} + 10^{-k} - 10^{-(k+1)}}{1 - 10^{-1}} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ which is what I wanted to prove.
Using the principle of finite induction, I now know that $$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ is true for all counting numbers, n.
But because $$\frac{9}{10} = 1 - 10^{-1}$$ this also proves that $$S_n = 1 - 10^{-n}$$ for all counting numbers, $$n$$.
Thus our lower bounds on what $$S$$ can be looks like this:
$$1 - 10^{-1} \, \lt \, 1 - 10^{-2} \, \lt \, 1 - 10^{-3} \, \lt \dots \lt \,1 - 10^{-(n+5)} \, \lt \, 1 - 10^{-(n+6)} \, \lt \, 1 - 10^{-(n+7)} \, \lt \dots \lt S$$
Now I want to prove that if $$X$$ is less than 1 then there are members of the family $$S_n$$ which are larger than it.
Say $$X \lt 1$$. Then $$0 \lt 1 - X$$. Then $$1 - X$$ is a
positive number. Then there is a real number $$\epsilon$$ such that $$e^{-\epsilon} = 1 - X$$ because $$f(x) = e^x$$ is a continuous function takes on all positive real values. Also $$f(x \ln 10) = e^{x \ln 10} = 10^x$$. So $$X = 1 - e^{-\epsilon} = 1 - 10^{- \frac{\epsilon}{\ln 10}} $$. So the question of if there are members of the family $$S_n$$ which are greater than $$X$$ reduces to the question of asking if there are counting numbers which are larger than the real number $$- \frac{\epsilon}{\ln 10} = - \frac{\ln ( 1 - X) }{\ln 10}$$ and the answer is yes, most of them.
So every number less than 1 is less than some (or all!) members of the family $$S_n$$ and thus less than $$S$$.
Thus 1 is greatest lower bound on what S could be, because all numbers less than 1 fail to qualify.
One of the biggest principles of the real numbers is that if a non-empty set of numbers has a upper (or lower!) bound then it has a least upper bound (or a greatest lower bound). By this, we see that the set of positive rational numbers whose square is less than 2 has greatest lower bound of 0 and a least upper bound of √2. 0 is not positive and √2 is not rational, but they are the minimal bounds of that particular set of positive rationals. That's why $$S$$ which is the least upper bound of the family of $$S_n$$ need not be a member of the family and can be 1.