Setup
Aside
Aside: This isn't relevant to the main thrust of your question, but
I wanted to note that there are many different equivalent definitions
of $0.\overline{9}$, depending on how exactly the real numbers are
set up. For instance, in some texts that expression would not be defined
as a summation of any kind, but rather would signify something like
"the unique real number in all of the intervals $\left[0,1\right]$,
$\left[0.9,1\right]$, $\left[0.99,1\right]$, ...".
For the rest of this answer, I will assume the summation definitions
$0.\overline{9}={\displaystyle \sum_{n=1}^{\infty}}\dfrac{9}{10^{n}}$
and $0.9_{N}={\displaystyle \sum_{n=1}^{N}}\dfrac{9}{10^{n}}$ where
$N\in{}^*\mathbb{N}\setminus\mathbb{N}$ is an infinite nonstandard
natural number.
Approach
I will be splitting this answer into three levels requiring the reader
to be comfortable with more "advanced" mathematics for each level,
but it is hard to target every person's mathematical background, so
please comment if some detail should be added somewhere.
Pre-calculus
$0.\overline{9}$ means the real number that $0.9$, $0.9+0.09=0.99$,
$0.9+0.09+0.009=0.999$ are getting closer to. As discussed in many
other places on the internet, including the English Wikipedia article
for 0.999..., the number it's getting closer to is $1$.
However, in some branches of mathematics, there are ways to imagine/define
new "numbers" that are not real numbers, which nonetheless can
be compared to real numbers. "Numbers" like $0.9_{N}$ are less
than $1$, but closer to it than any real number less than $1$ could
be.
Calculus
A definition of $0.\overline{9}$
In an expression like ${\displaystyle \sum_{n=1}^{\infty}}\dfrac{9}{10^{n}}$,
the $\infty$ symbol on the summation sign is a shorthand for evaluating
a certain limit. Specifically, we evaluate the limit of the sequence
of finite partial sums: ${\displaystyle \lim_{m\to\infty}}\left({\displaystyle \sum_{n=1}^{m}}\dfrac{9}{10^{n}}\right)$.
And here, this $\infty$ symbol means something that can be explained
using only everyday real numbers, which are always finite. Specifically,
this limit is a number $L$ so that for every error tolerance $\varepsilon>0$,
we can choose a natural number $M$ big enough so that if $m\ge M$,
the finite sum is "$\varepsilon$-close to $L$" - i.e. that $\left|{\displaystyle \sum_{n=1}^{m}}\dfrac{9}{10^{n}}-L\right|<\varepsilon$.
We know from our experience with geometric series that the limit $L$
should be $\dfrac{9/10}{1-(1/10)}=1$ in this case. And this works out with the $\varepsilon$
definition. For example, if $\varepsilon=0.003$, then we can choose
$M=3$ since $\left|{\displaystyle \sum_{n=1}^{3}}\dfrac{9}{10^{n}}-1\right|=\left|0.999-1\right|=0.001<0.003=\varepsilon$
and $\left|{\displaystyle \sum_{n=1}^{4}}\dfrac{9}{10^{n}}-1\right|=\left|0.9999-1\right|=0.0001<0.003=\varepsilon$,
etc.
(In fact, we can handle any $\varepsilon$ by taking $M$ to
be the ceiling of $\log_{10}\dfrac{1}{\varepsilon}$.)
I emphasize again that the above discussion did not involve any infinite
numbers, just a symbol $\infty$ which was a directive to do certain
work with finite inequalities.
The idea of $0.9_{M}$
Without getting into the weeds of details, there are ways to add on
new "numbers" to the naturals and to the reals which get the name
"nonstandard". For instance, there could be an infinitesimal "number"
$c$ that's smaller than all of $0.1$, $0.01$, $0.001$, ..., but
still greater than $0$ (no real number is like that). It turns out
that some numbers like $1-c$ where $c$ is a certain kind of infinitesimal
might be deserving of a name and symbols that evoke a similar idea
to the summation of $0.0\ldots09$s we did earlier. But those are different
"numbers" than $1$, the only real number close to those finite
sums.
Just like summation had a special unintuitive meaning when the symbol
on top was $\infty$, it has a different special unintuitive
meaning when the symbol on top is "an infinite nonstandard natural
number", whatever that is. The precise meaning of ${\displaystyle \sum_{n=1}^{N}}\dfrac{9}{10^{n}}$ is not only different than ${\displaystyle \sum_{n=1}^{\infty}}\dfrac{9}{10^{n}}$, but also depends on which system of nonstandard numbers we're working with, and which "infinite nonstandard natural" number $N$ is
within that system.
You can learn about this on an informal level in Keisler's free textbook Elementary Calculus: An Infinitesimal Approach.
Analysis
To really compare these two expressions, one of which only makes sense
in a context of non-standard analysis, it's best to view each of them
through that same lens. This will take a bit of development/statements
of basic facts about non-standard analysis, but I'll try to thread
the needle in only being detailed enough to address the original question,
rather than write a proper development of nonstandard analysis.
Hyperreals
Probably the most hands-on approach to constructing "nonstandard"
numbers is to create Robinson's Hyperreals
via an ultrapower construction.
Quoting my other answers about hyperreals here:
For those unfamiliar, the basic idea behind of the construction isn't
too complicated. I like Terry Tao's voting
analogy.
A hyperreal is a sequence of reals (well, an equivalence
class of sequences)
that vote each time you ask about a property (like "are
you bigger than $5$?" ). How to determine which infinite
collections of voters count as good majorities is handled by some
technical stuff, but you don't have to worry about that to get the
idea.
One important property (which would require more details to make precise
and prove) is that essentially any functions/operations that you can apply
to reals (or a relevant subset, like the naturals), can be applied
to these hyperreals: just do everything entry
by entry (for any representative of the equivalence class). For example,
$x=\left[\left(1,2,3,\ldots\right)\right]$ is an "infinite nonstandard
natural" (since a majority of the entries would agree that it's
greater than $17$, etc.). And if you want to square it, just square
each entry to get $x^{2}=\left[\left(1,4,9,\ldots\right)\right]$.
Since $4>2$, $9>3$, etc., we have $x^{2}>x$.
Another key property is that if a hyperreal is finite (a majority
of the entries are bounded between two finite reals), then there is
a closest real number to it. This is sometimes called the "standard
part" or the "shadow" of the original hyperreal.
What is $0.9_{N}$?
In order to interpret something like "$0.9_{N}={\displaystyle \sum_{n=1}^{N}}\dfrac{9}{10^{n}}$
where $N\in{}^*\mathbb{N}\setminus\mathbb{N}$ is an infinite nonstandard
natural number.", we need to define finite summations, and use that
to understand what this should mean. For instance, there is a finite
sum function $f:\mathbb{N}\to\mathbb{R}$ given by $f\left(m\right)={\displaystyle \sum_{n=1}^{m}}\dfrac{9}{10^{n}}$.
And an infinite nonstandard natural is like a sequence where all (or
at least a "majority") of the entries are naturals, and that no
majority of the entries are bounded. $x$ and $x^{2}$ from the previous
paragraph are examples of these. So given $N=x^{2}=\left[\left(1,4,9,\ldots\right)\right]$,
we can define the sum $0.9_{N}={\displaystyle \sum_{n=1}^{N}}\dfrac{9}{10^{n}}$
to be $f\left(N\right)$, which is $\left[\left(0.9,0.9999,0.999\,999\,999,\ldots\right)\right]$.
If we instead had $N=x=\left[\left(1,2,3,\ldots\right)\right]$, then
$f\left(N\right)$ would be $\left[\left(0.9,0.99,0.999,\ldots\right)\right]$,
which is a smaller number, since $0.99<0.9999$ and $0.999<0.999\,999\,999$,
etc. But all of these possibilities for various values of $N$ are
still less than $1=\left[\left(1,1,1,\ldots\right)\right]$, since
$0.9\ldots9<1$ for any finite amount of $9$s you put there. For
this reason, something like $0.9_{N}$ is different from $1$.
Limits via Hyperreals
A reason to develop something like the hyperreals is that they give
you a way to develop calculus/analysis that more closely resembles
intuition about infinitesimal quantities held by Fermat and Leibniz.
There are translations of definitions of limits, derivatives, integrals,
etc. into discussions of infinite/infinitesimal hyperreals.
For our purposes, we just need to worry about the limit of a sequence.
If $\left(a_{1},a_{2},\ldots\right)$ is a sequence, then the function
$a\left(m\right)=a_{m}$ is a function on the positive integers, which
we could apply to infinite nonstandard naturals like $x$ or $x^{2}$.
If the closest real number to $a\left(N\right)$ is $L$ no matter
what infinite nonstandard natural we pick, then we say that ${\displaystyle \lim_{n\to\infty}}a_{n}=L$.
It turns out that this agrees with the $\varepsilon$ definition discussed
earlier.
This means that ${\displaystyle \sum_{n=1}^{\infty}}\dfrac{9}{10^{n}}={\displaystyle \lim_{m\to\infty}}{\displaystyle \sum_{n=1}^{m}}\dfrac{9}{10^{n}}$
is basically saying "what is the closest real to any of the possibilities
for $0.9_{N}$?". And the answer to that is just $1=0.\overline{9}$.
Conclusion
$0.9_{N}$ is some weird hyperreal/nonstandard real which is infinitesimally close to $1$, leveraging that $N$ is a particular infinite quantity. But the use of the $\infty$ symbol means that we are no longer working with a particular infinite quantity, just asking which everyday real number all of those $0.9_{N}$s are close to, which is $1$.
"Infinite $N$"
This question hinged on "$\infty$" and "infinite $N$" both pointing to a concept of infinity, and the distinction between them. There are actually very many different concepts of infinity, and it can be challenging and rewarding to keep the distinctions and connections straight. You can read summaries of them in my answer to the MathSE question Understanding infinity.
