Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
Solution 1:
First let me tell you that the idea that an infinite sequence "ends with something" is a solid idea. It's a perfectly natural one. The point is that the sequence is not indexed by $\Bbb N$, anymore, but rather by $\Bbb N\cup\{\infty\}$, where $\infty$ is another point, which lies after all the natural numbers.
The point is that an "infinite sequence" is a very general notion. People just often like to think about sequences which are indexed only using the natural numbers (with their natural ordering, that is). But as you will progress in your studies you might meet other objects which are indexed using other infinite sets.
And the reason people often limit themselves to sequences indexed by the natural numbers is that for the real numbers (and similar concepts), these sequences are enough. In this case, of the real numbers, we have that each real number can be defined as a limit of decimal digits, as others have explained, and therefore $1.\underbrace{000\ldots}_{\text{infinite }0\text{'s}}1$ is not a definition of a real number.
Note that this is not the limit of $1+(\frac1{10})^n$, either, which colloquially might be written as $1+(\frac1{10})^\infty$. That limit would be the limit of $1.1,1.01,1.001,\ldots$ and you can see that at no point in this sequence there is a number with infinitely many $0$'s written after it. And indeed this limit would be equal to $1$.
This is also different from the $0.999\ldots$ situation, since it is a sequence indexed by $\Bbb N$, which can be seen as the limit of its initial segments. Whereas a sequence indexed by $\Bbb N\cup\{\infty\}$ is not the limit of its initial segments, since none of them include information about the last digit.
So does it exist? Yes. It's just not a real number. It's a sequence of digits indexed by something other than $\Bbb N$.
Finally, Let me point out that as far as the concept of infinity goes in calculus, it's not quite unique. There is one infinity which signifies arbitrarily large values, another which signifies arbitrarily large negative values, there are infinities which ignore the sign at all, when you talk about a smooth function that can be differentiated infinitely many times, the infinity here is in fact "infinite sequence" rather than the infinities mentioned before, and it's a completely different type of infinity.
And there are other infinities which you might encounter, even in a calculus class.
Solution 2:
The bad news:
That string of symbols $$ 1.0\cdots01 $$ has no meaning as a real number. Meaning or rather semantics would be a valid mapping from an infinite string $s$, e.g. from $\Sigma^\omega$ (link), to $\mathbb{R}$.
The string representation of a floating point real number in base $10$ by convention means e.g. $$ (1.0\cdots)_{10} = (\underbrace{d_0.d_1 d_2 d_3 \cdots }_{\mbox{string}})_{10} = \underbrace{\sum_{k=0}^\infty d_k 10^{-k}}_{\mbox{real number}} $$ Where would you like to add that last $1$ digit?
The good news:
It might be something else.
Of course you can define your own semantics and operations on the strings but you will most likely end up with something that behaves more or less differently from the real numbers (like finite IEEE floats and their operations are not the same as the full set of real numbers and the basic operations on it, but something very close).
And the ugly news: $\tiny \mbox{(c) by Asaf Karagila}$
As fellow user Hyrkel noticed, that finite string I used $$ [1] [.] [0] [.] [.] [.] [0] [1] $$ to suggest an infinite string with two ends (prefix "$1.$" and suffix "$1$") and infinite many $0$ symbols in between is problematic too - interpreted as a string already.
Is this a proper string? is it even a proper infinite string? How would you be able to recognise it?
Next step would be to attach a meaning to it, preferably some number, but I will only argue on the above questions which are not simple already.
Computer scientists stick to mathematical machine models, like the finite automaton or the Büchi automaton to reason about strings. These machines can either accept or reject a string they are presented with. Their recognition process resembles the process of a sensor reading a linear tape or track from left to right. Even the variants for infinite strings act like this.
The infinity here is not so much problematic because of it's sheer size but rather because of it's dullness: what reason should the automaton have to stop the recognition of the infinite part and proceed with the finite suffix? The recognizable infinite strings seem to be of the variant one end finite, one end infinite. (Do not nail me on this)
I am not sure if a non-deterministic (multiple choices possible) Büchi automaton that would accept the string $1.0^\omega$ could be properly extended to recognise $1.0^\omega 1$.
I would attempt it by adding another arc from the final state to itself which is accepting the final $1$ symbol. That would work to accept $1.0^\omega1$ but it would also still accept just $1.0^\omega$. That makes it not much useful, what I can not distinguish is practically the same.
The solution is probably another automaton that starts recognition simultaneously from both ends or some mapping which lists the infinite sequence alternating from both sides at once, something like $$ (1 1) \, (. 0) \, (0 0) \cdots $$ this would resort to established structures but I am not aware of such approaches.
Solution 3:
In real analysis, $$\lim_{n\rightarrow\infty}\left( 1+\frac{1}{10^n}\right )=1$$ Even if it does not mean anything to say "infinite number of zeros" in real analysis, we can suppose this number is equal to $1$.
But in the field of surreal numbers, it's not the same. This number exists and will be equals to $1+\frac{1}{\omega}$, if you consider the infinity number of zeros to be $\omega$.
Solution 4:
If a sequence $1.00000\dots $ is infinite it can't have an end $\cdots 0001$. Infinite means endless.
A finite sequence is a line of (mathematical) objects $a_0,a_1,a_2,\dots, a_n$. But it seems to be some disagreement what an infinite sequence is. At least I disagree.
Obviously, the object $a_k$ represent a function $k\mapsto a_k$ with ordered indices $k$, but could it be any function? Due to Wikipedia:
Most precisely, a sequence can be defined as a function whose domain is a countable totally ordered set, such as the natural numbers.
In that case a function $\mathbb N\cup\{\infty\}\rightarrow A$ (for some set $A$) is a sequence $a_0,a_1,a_2,\dots$ with a last element $a_\infty$, without an immediately preceding element in the sequence.
In my intuition and in my opinion any element in a sequence, except the first, has an immediately preceding element.
Latin: sequentia (“a following”).
However, it's possible to generalize to "bi-sequences" $(S_1,S_2)$ when $S_1=(1,0,0,\dots)$ is the initial sequence and $S_2=(\dots,0,0,1)$ is the termimal sequence, and define an arithmetic for "numbers" defined by bi-sequences as $(S_1,S_2)$.
Solution 5:
In the real number system $\mathbb R$, there is no such number.
P.S. There is no such number in the complex numbers $\mathbb C$, either.
P.P.S. Sorry.