Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $\zeta(-1)$. You've outlined using the functional equation to relate it to $\zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3\zeta(-1)=\eta(-1)=\lim_{x\to 1^-}\left(1-2x+3x^2-4x^3+\cdots\right)=\lim_{x\to 1^-}\frac{1}{(1+x)^2}=\frac14$$
where $\eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+\mathrm{\&c}=\frac{1}{(1+1)^2}=\frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $\zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $\zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $\zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
\begin{alignat}{7}
\zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+\cdots& \\
2\times2^{-s}\zeta(s)&{}={}& 2\times2^{-s}&& {}+2\times4^{-s}&&{} +2\times6^{-s}+\cdots& \\
\left(1-2^{1-s}\right)\zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+\cdots&=\eta(s) \\
\end{alignat}
$$
The identity $(1-2^{1-s})\zeta(s)=\eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3\zeta(−1) = \eta(−1)$. Now, computing $\eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $\zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
this had been a comment, but is now meant as an answer introducing the citation from E. Delabaere, Université d' Angers
I've just skimmed the intro of the Candelspergher-book, and have not much time to go deeper into it. But I see that he says, that the notation $\qquad \displaystyle \sum_{n \ge 0}^\mathcal R \cdots \qquad$ means to have captured the pole of the zeta.
As far as I've understood this, this means that the singularity of the $\zeta(1)$ is removed - and this result is called "Ramanujan sum".
So what he calls the "Ramanujan sum" is actually $\zeta(s)-1/(s-1)$. It seems that it is perhaps a unlucky misnomer. Possibly it were better (like with the "incomplete gamma-function") to write
"The Ramanujan sum of the zeta is the incomplete zeta" or the like,
and thus this should then be called "Ramanujan incomplete sum" to indicate that a completing-term is systematically missing from the sum of the series under discussion. The including of the completion-term would then be called with the common name "Ramanujan-summation"
Then there would be nothing irritating when writing
The "Ramanujan incomplete sum" of the series $1+2+3+4+...$ is $$\sum_{n \ge 1}^{\mathcal R} n = \zeta(-1)-\frac1{-1-1} = -\frac1{12} + \frac12 = \frac5{12}$$
and must be completed by $ - \frac12 $ to arrive at the known value $ - \frac1{12} $ for the zeta-interpretation of this series.
Just my 2 cents...
update for completeness of my arguments I just include a snippet from E.Delabaeres article on "Ramanujan summation" by the summary of Vincent Puyhaubert, page 86.
- Legend: Here $a(x)$ are the terms of the series, rewritten as when the full series $a_1+a_2+a_3+...$ is expressed in the transformed form $a(1)+a(2)+ a(3)+\cdots $ and the powerseries-representation of $a(x)$ is combined with the Bernoully-numbers (according to the Euler-Maclaurin-formula for this problem)
- The background-colored elements and red ellipses are added by me for pointing to the important terms-of-formula
Best Answer
There are answers in -as I think- in mathoverflow for "what are physical interpretations of zeta at negative arguments" (or the like, I don't have the actual words in mind). Just browse a bit through MO or MSE using that terms as filter.
0) Need of a new concept - and a basic requirement: Make the general term explicite
For the interpretation and assignment of a meaningful value in your series you must define, how the general term, say $a_k$, shall be constructed, i.e. how the value of $a_k$ is dependend on its index $k$. This is a requirement which shall be demanded in all discussions of divergent series (and should not been forgot when such series, written only in their "obvious" numerical forms, are considered at all).
In this example the value of the general term, say $a_k$, reduces simply to the index $k$ if taken from $k=1 \ldots \infty $ and we have actually the formal statement $S = \sum_{k=1}^\infty k$. But because the summing procedure is obviously divergent we must introduce some new concept if we want assign something meaningful to it.
1) Geometric series
A useful idea here was (for instance already by L. Euler) that we can cofactor a variable $x$ at the terms $a_k$ such that for some $x$ the series becomes convergent and evaluatable to some finite value and see what happens with the sequence of resulting sum-values, when the expressions of $x$ approximate $1$ . For instance, we could redefine your sum as $$ S(x) = 1 + 2x + 3x^2 + 4x^3 + ... \\ S = \lim_{x \to 1^-} S(x) $$ because there are some $x$ (actually it is important, that there exists a continuous interval of $x$) where this is convergent, for instance for $x=0 \ldots \frac12$. We can then even observe that with that definition the sum $S(x)$ has a closed form $$S(x) = {d\over dx}{1\over 1-x } = {1\over (1-x)^2 }$$ This is already a very nice representation, because it allows now even negative $x$ which result in finite values for the alternating sum, even if divergent, because it is accepted to assign that fraction's value also to its formal power series-expression even if the latter is divergent - except, well, except if $x=1$.
But what we want here is actually just the latter, so we have not yet a satisfying answer.
2) Dirichlet series
Another idea is to apply a function of $x$ to the exponent of the terms $a_k$ and then let $x$ go to $1$ . Thus we consider the notation $$ S(x) = 1^x + 2^x + 3^x +4^x + ... $$ which is convergent for a continuous interval of $x$, for instance of $x = -\infty \ldots -2$ (and even a bit more). But still, for $x=1$ we get no obvious answer.
But interestingly, for the whole interval of convergence we have also the functional relation $$ \begin{matrix} \text{let } &A(x) &=& 1^x - 2^x + 3^x - 4^x + \ldots \\ \text{then } &S(x) &=& A(x) + 2 \cdot 2^x \cdot S(x) \\ \text{and} & S(x) &=& A(x)/(1-2 \cdot 2^x) \end{matrix} $$ And now we can assign a meaningful value to $S = \lim_{x \to 1^- }$. Either by evaluating $A(1)$ as conditionally converging series in the given form, or by the above geometric-series interpretation and its derivative at $x=-1$ where both ways of evaluation give the same rational expression $A(1)=\frac 14$.
After that nothing more divergent is there and we get $$ S = \lim_{x \to 1^-} S(x) = \frac 14 / (1-2\cdot 2^x) = \frac 14 / (1-4)= - \frac 1{12} $$ having now a proposal for a meaningful assignment of a finite value for the infinite divergent series.
3) Caveat
Of course that value must conform with all and any cases where such series occur in mathematics, and one example of a contradiction with known results taken from the mathematic without divergent summation would invalidate that procedure!
Moreover, we would hope that even in the physical world, where we model some observations with that series, such an assignement of values to a divergent series would hold. Interestingly such observations exist in the real world and it seems that the whole process and also the final value meets the modeling of that observations. (Examples are given to the according questions either in MSE here or in mathoverflow (I've just found (1)(2)), you can do a search for the important words "zeta" and "at negative arguments" or similar)