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Is (2/infinity = 1/infinity)?
Essentially, you gave the answer yourself: "infinity over infinity" is not defined just because it should be the result of limiting processes of different nature. I.e., since such a definition would be given for the sake of completeness and coherence with the fact "the limiting ratio is the ratio of the limits", your
and, say (this is my choice)
would have to be equal (as they commonly define 
), which does not happen.
I will quote the following from Prime obsession by John Derbyshire, to answer your question.
Nonmathematical people sometimes ask me, “You know math, huh? Tell me something I’ve always wondered, What is infinity divided by infinity?” I can only reply, “The words you just uttered do not make sense. That was not a mathematical sentence. You spoke of ‘infinity’ as if it were a number. It’s not. You may as well ask, ‘What is truth divided by beauty?’ I have no clue. I only know how to divide numbers. ‘Infinity,’ ‘truth,’ ‘beauty’—those are not numbers.”
Every claim you started with is wrong. Your $S_{1}, S_{2}, S_{3}$ are all divergent series.
It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value.
You start with wrong conditions.
For example, the infinite sum $\sum_n^{\infty} (-1)^n $ is not equal to $1/2$.
I object to your teacher's answer (even if we remove physical obstructions, such as indivisibility of some particle or another): after any finite amount of time, Aladdin will have only divided it into pieces of size $(1/2)^n$ for some finite $n$; this is still positive. However, the limit is zero, which is what is meant by $(1/2)^\infty$. As $n$ gets bigger, $(1/2)^n$ gets as small as you want, so we say that its limit as $n \rightarrow \infty$ is zero.
What your teacher said was not very mathematically exact. You have to observe the infinite sequence of powers of one half,
$$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots$$
And the proper term is that the sequence tends to zero (it converges to its limit which is $0$).
The fact that this goes to zero shouldn't surprise you. But you need an infinite number of cuts.
Of course this is assuming that he's throwing the rest of the carpet away. Otherwise you just get a heap of differently sized pieces :)
It doesn't. there's a function called the reimann zeta function. It has a complicated definition but $\zeta (s) $ will equal $1/1^s + 1/2^s + 1/3^s+... $ IF that expression has a value. If that expression does not have a value $\zeta (s) $ will have a different value.
As it turns out $\zeta (-1) = -1/12$. Now IF 1+2+3+... converged (it doesn't obviously but if it did) then it would have to be that 1+2+3+... = $\zeta (-1)=-1/12$. But 1+2+3+... DOESN'T converge so this is utterly irrelevant and meaningless.
Here's an analogy. It's a different result but it's a similar idea:
Let $N_x = 1 + x + x^2 + x^3 + ....$. To cut to the chase, if $-1 < x < 1$ then $N_x = \frac 1{1 - x}$. This is because $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$.
So, for example $N_{1/2} = 1 + 1/2 + 1/4 + 1/8 + .... = 2 = \frac 1{1 - 1/2}$. And $N_{-1/2} = 1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + ... = \frac 1{1-(-1/2)} = \frac 2 3$.
Neat, huh?
But does that mean $N_{-1} = 1 -1 + 1 - 1 + 1 - 1 + .... = \frac 1{1-(-1)} = 1/2$? Or that $N_{2} = 1 + 2^2 + 2^3 + 2^4 + .... = \frac 1{1-2} = -1$?
Obviously not. Why not? Well, because when we said $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$, we were assuming $(1 + x + x^2 + x^3 + ...)$ converges to a meaningful value. It does converge to a meaningful value if $-1 < x < 1$ and if so then everything we said was true. But if $|x| \ge 1$ then $(1 + x + x^2 + x^3 + ...)$ doesn't converge to a meaningful value and nothing we said makes any sense.
So it's the same thing with $1 + 2 + 3 + 4 + ....$. IF $1 + 2^{-s} + 3^{-s} + ....$ equals anything than that thing equals $\zeta(s)$. But $1 + 2^1 + 3^1 + ...$ doesn't equal anything. So it doesn't equal $\zeta(-1) = -1/12$.
The "usual" sum of this series is $\infty$, since $$ \sum_{j=1}^\infty j = \lim_{m \to \infty} \sum_{j=1}^m j = \lim_{m \to \infty} \frac{m(m+1)}{2} = \infty $$ Your friend is talking about Ramanujan summation, which gives finite values to some divergent series. Don't get mislead by Numberphile!
The Ramanujan sum of $\sum_{j=1}^\infty j$ corresponds to $\zeta(-1)$, where $\zeta$ is the Riemann $\zeta$ function $$ \zeta(z) = \sum_{j=1}^\infty \frac{1}{z^j} \qquad (\Re (z) > 1) $$ analytically continued to the entire complex plane except for $1$.