I think I understand intuitively how we can assign a value to the sum of all natural numbers. But of all the proofs that I've seen that show why $\zeta(-1) = -\frac{1}{12}$, none of them use their own tactics to address the divergence of the harmonic sum.
To give an example of what I mean, take a look at one of the most famous proofs for the sum of the natural numbers:
$$
c = 1 + 2 + 3 + 4 + … \\
4c = 4 + 8 + 12 + 16 + … \\
-3c = 1 – 2 + 3 – 4 + 5 – … \\
1 – 2 + 3 – 4 + … = \left.\frac{1}{(1 + x)^2}\right|_{x=1} = \frac{1}{4} \\
-3c = \frac{1}{4} \\
c = -\frac{1}{12}
$$
If you were considering finding a value for $c$ whilst keeping in mind convergence, you would stop immediately at the first line of that proof. Clearly, $c = \infty$, no doubt about it. Yet if we continue on with the proof, using algebraic manipulations that would seem absurd to anyone considering convergence, we end up with $c = -\frac{1}{12}$.
So obviously we don't mean that the sum of the natural numbers converges to $-\frac{1}{12}$, but rather we mean that if $c$ were to be some number, it would have to satisfy the given algebraic properties ($-3c = \frac{1}{4}$).
This idea of analytic continuation makes sense to me, but what doesn't make sense to me is why it can't be applied to the harmonic sum. If we completed disregarded the convergence of $\sum_{k=1}^\infty{\frac{1}{k}}$ like we did above, are there algebraic manipulations we could apply to the sum that allow us to assign a value to it? Why can't we just define $H = \sum_{k=1}^\infty{\frac{1}{k}}$ to be a number that has certain algebraic properties, like we did with the sum of the natural numbers? Can you prove that $H$ must be $\infty$ using the same process as above?
Best Answer
Letting $$ H=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots $$ and assuming convergence, we have $$ 0=H-2\cdot\frac{1}{2}H=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right) - 2\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)=\\1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\ldots=\ln 2. $$ Since this is a contradiction, the sum must diverge.
Update.
To be more clear about what I'm saying here... I'm not saying that there's no way to assign a value to the harmonic series. Clearly there are any number of ways, some sillier and more arbitrary than others. Call a partial function $S$ from infinite sequences to real or complex numbers a summation method if $S(\{a_i\})$ is defined and equal to $\sum_{i=1}^{\infty}a_i$ whenever the sum converges (in the usual sense); and say that $\{a_i\}$ is $S$-summable if $S(\{a_i\})$ is defined. There are a number of properties that you might want your summation method to preserve from ordinary summation. For instance, you might want it to remain linear: $$S(\{ \alpha a_i + \beta b_i \})=\alpha S(\{a_i\}) + \beta S(\{b_i\})$$ whenever $\{a_i\}$ and $\{b_i\}$ are $S$-summable. You might also want it to remain stable under insertion of zeros: if $J:\mathbb{N}\rightarrow\mathbb{N}$ is increasing and $\{a_i\}$ is $S$-summable, then $S(\{a'_i\})=S(\{a_i\})$, where $a'_{J(i)}=a_i$ and $a'_i=0$ for $i\not\in J^{-1}(\mathbb{N})$. I haven't said anything about analytic continuation or whatever other exotic regularization you want to consider. But what I'm saying, above and in the comments, is that no summation method under which either the harmonic series or $1+2+3+\ldots$ is summable can be both linear and stable.
Now, there's a weaker form of stability (under insertion of only finitely many zeros) that you can preserve while making the harmonic series summable, but not while making $1+2+3+\ldots$ summable. So in that specific sense, the latter series (!) is more pathological. In order to sum the natural numbers, you need to give up either linearity or stability, rendering the manipulations in the original post meaningless.