TITLE: Using the rules that prove the sum of all natural numbers is $-\frac{1}{12}$, how can you prove that the harmonic series diverges?
QUESTION [13 upvotes]: 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?

REPLY [2 votes]: For anyone who happens upon this question and doesn't understand, like I did, I would like to elaborate on exactly what I specifically was missing in my understanding.
When I posted this question, I didn't know what people meant when they said $\sum_{n=1}^\infty n = -1/12$. Now that I'm older and a bit more experienced, this statement is mathematically wrong. Equating these two quantities is simply a false, pseudo-mathematical inanity. No one told me this when I heard about it, so I always perceived it as mathematical fact.
So what do people actually mean when they write that? Well, we can define the Riemann Zeta Function:
$$
\zeta(s) = \sum_{n=1}^\infty\frac1{n^s}
$$
for a complex number $s$. We know that this sum converges for $\Re(s)>1$, and it diverges when $s=1$, and $\Re(s)<1$. However, quite frequently we talk about values of the zeta function for which $\Re(s)<1$. This is because we're not really talking about the zeta function, but rather an (and the unique) analytic continuation of the zeta function.
To be specific, if a function $f$ is analytic in a domain $U$, and $V\supset U$, and $F$ is some function analytic on $V$ such that $F(z)=f(z)$ for $z\in U$, then $F$ is an analytic continuation of $f$ to a larger domain. It turns out that this function $F$ is unique. Since the zeta function is analytic in the half-plane $\Re(s)\ge 1$, minus the point $s = 1$, it has an analytic continuation to $\mathbb{C}\backslash\{1\}$, which is uniquely determined by the behaviour of the function $\zeta(s)$ in the domain of convergence of the infinite sum.
One may be curious as to why this analytic continuation cannot be "infinite everywhere" (which would seem to be the only logical choice for such a function given that the sum doesn't converge anywhere outside $\Re(s)>1$). If this were true, then the reciprocal of that analytic continuation would be identically zero on a dense subset of the complex plane, which, by the principle of analytic continuation once again, must imply that the function is identically zero everywhere. But this doesn't make sense since $1/\zeta(s)$ is clearly defined for $\Re(s)>1$.
Hence, there is some meaning to the values $\zeta(s)$ for $\Re(s)<1$, but this does not mean that $\zeta(s) = \sum_{n=1}^\infty\frac1{n^s}$ for all $s$. The sum diverges. The zeta function is extended.
However..., if we abuse our notation a bit, and let $\sum_{n=1}^\infty\frac1{n^s}$ denote the analytic continuation, rather than the actual sum, for $s$ outside the domain of convergence, then we can plug in $s = 1$ to get
$$
\zeta(-1) "=" 1 + 2 + 3 + 4 + \cdots
$$
and one can prove, properly (i.e. using real math, not pseudo-math), that $\zeta(-1)$ (that is, the analytic continuation of $\zeta$ evaluated at $s=-1$), is indeed equal to $-1/12$.

As for the sum $\sum_{n=1}^\infty\frac1{n}$, this is simply a pole of the Riemann zeta function. The zeta function, on the entire real line, is a meromorphic function (i.e. it is holomorphic/analytic everywhere except on a countable set with no limit points). Functions like these are "allowed" to have singularities on a countable number of disjoint points. They can be thought of as parts where the function "looks like" $1/(s-z)^k$ for some $k$. (There are also "essential singularities" which go to infinity faster than this for any $k$, but the pole of the zeta function is not essential).
Essentially, what this means is that $\sum_{n=1}^\infty n = -1/12$ doesn't contradict some notion of convergence/divergence which implies $\sum_{n=1}^\infty\frac1n = \infty$. Both diverge, we just have a "special way" of evaluating the former sum.