Translations of my "summary" into English.

For any counting number, $$n$$, the expression $$S_n$$ is just a shorthand way of writing the quantity $$1 - 10^{-n}$$ and this follows from the original definition of $$S_n$$ being the finitely repeating decimal 0.9...9 with $$n$$ 9's, or the same value expressed as a geometric sum which starts with 0.9 and each term is smaller by a factor of 10, or the clever formula for writing in fixed form the sum of any finite geometric series. So all of these terms are exactly equal to each other even though they don't look much alike:

$$S_n = 0.\underbrace{999...999}_{n \; \textrm{nines}} = \sum_{k=1}^n \frac{9}{10^k} = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}} = 1 - 10^{-n}$$

$$ T_n = 9 \times 10^{-n} $$

For any counting number, $$n$$, the expression $$T_n$$ is just a shorthand way of writing the quantity $$9 \times 10^{-n}$$ and since $$S_n = \sum_{k=1}^n T_k$$ it follows that $$T_{n+1} = S_{n+1} - S_n$$ and also $$S_n + \frac{1}{9} T_n = 1$$

$$\lim_{n\to\infty} S_n = S = 0.999... $$

As the parameter $$n$$ is moved to larger values, the expression $$S_n$$ moves ever closer to a unique limiting value, which we $$S$$. Conceptually, this value must be the same as if the number of repeating nines grew larger than any finite number, so we choose to also call it $$0.999...$$ which is the topic of this thread. Even though the symbol $$\infty$$ is in this expression we do not use that symbol in any formal sense. We don't set the parameter $$n$$ to infinity because that would lead us to an area where arithmetic is not defined.

$$\lim_{n\to\infty} T_n = T $$

Likewise, as the parameter $$n$$ is moved to larger values, the expression $$T_n$$ moves ever closer to a unique limiting value which we call $$T$$. If we wanted we could call that $$T = 9 \times ( 1 - 0.999... )$$ but since we don't yet know how to do the arithmetic in the inner parenthesis, I submit introducing this notation is a bad idea. We will attempt to reason out what $$0.999...$$ must be and that should solve the problem of what $$T$$ is at the same time.

$$\forall n \in \mathbb{N}^+ \quad S_n + \frac{1}{9} T_n = 1 \; \wedge \; S_n \lt S \; \wedge \; T_n \gt T $$

For any specific counting number $$n$$, all of the following are true: $$\quad S_n + \frac{1}{9} T_n = 1$$ which follows from direct calculation and $$S_n \lt S$$ and $$T_n \gt T$$ which both follow from the observation that the expression $$10^{-n}$$ is strictly decreasing for increasing values of the parameter $$n$$.

$$S \not\in \left{ S_n \; | \; n \in \mathbb{N}^+ \right} $$

The value $$S$$ is not anywhere equal to the expression $$S_n$$ for any counting number $$n$$. That makes sense since $$0.999...$$ with a greater than finite number of 9's can't equal $$0.\underbrace{999...999}_{n \; \textrm{nines}}$$ when $$n$$ is finite.

$$ T \not\in \left{ T_n \; | \; n \in \mathbb{N}^+ \right} $$

The value $$S$$ is not anywhere equal to the expression $$S_n$$ for any counting number $$n$$.

$$ x \lt S \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( 1 - x ) \; \rightarrow \; x \lt S_n \lt S $$

If $$x$$ is any real number less than $$S$$ then it follows that every counting number, $$n$$, which is larger than $$- \log_{10} ( 1 - x )$$ it follows that $$S_n$$ is strictly between $$x$$ and $$S$$.

From the above, it is impossible that $$S \lt 1$$ or there would be values of $$S_n$$ larger than $$S$$ while if $$S \gt 1$$ then there would be values of $$x$$ between the largest possible values of $$S_n$$ and $$S$$. Thus $$S = 1$$ and $$0.999... = 1$$.

$$ x \gt T \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( x / 9 ) \; \rightarrow \; T \lt T_n \lt x $$

If $$x$$ is any real number greater than $$T$$ then it follows that every counting number, $$n$$, which is larger than $$- \log_{10} ( x /9 )$$ it follows that $$T_n$$ is strictly between $$T$$ and $$x$$.

This should not be a surprise and it also follows from $$T = 9 \times ( 1 - 0.999... ) = 9 \times ( 1 - 1 ) = 9 \times 0 = 0$$

$$ \forall k \in \mathbb{N}^+ \quad \textrm{card} \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad = \quad k $$

If $$k$$ is any counting number, the size of the set of counting numbers not greater than $$k$$ is equal to $$k$$. Think about it. The set of the counting number not greater than 5 is {1, 2, 3, 4, 5} and the size of that set is 5.

$$ \forall k \in \mathbb{N}^+ \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad \prec \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; n \gt k \right} $$

If $$k$$ is any counting number, the set of counting numbers greater than $$k$$ strictly dominates the set of counting numbers not greater than $$k$$. Or: for any counting number, $$k$$ most counting numbers are larger than it.

$$ \infty \not\in \mathbb{N}^+ $$

Infinity is not a counting number. In many number systems it is not a full-fledged number that obeys the axioms of all other numbers.