number theory - Meaning of equality in zeta regularization

It is known that
$$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$
but it is also known that
$$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$
which can obtained using the analytic continuation of $\zeta(s)$. My question is: What is the true meaning of equality here? It is a common practice to write $f = {\mathcal O}(g)$ to mean $f \in {\mathcal O}(g)$. Is this something similar?

Answer

One thing that is sometimes lost when talking about series is what the symbol $$\tag{1}\sum_{n=1}^\infty a_n$$ means.

The point is that it is defined as a limit, i.e. the usual definition of the symbol $(1)$ is that it is the limit (if it exists) of the sequence of partial sums. This is already subtle, because the notion of limit depends on the topology. The topology is canonical on $\mathbb R$ or $\mathbb C$, but not in normed vector spaces where one can still consider series. This definition also stumbles with the case of conditionally convergent series, where any other limit can be achieved by reordering.

Anyway, the definition of $(1)$ as the limit of the sequence of partial sums (which is $\infty$ in your example) is not the only possible definition. There are other notions, such as Cesàro-sum and Abel-sum (these two the most noteworthy among others).

Yet another meaning one can given to the $\sum\limits_{n=1}^\infty n$ is the analytic continuation you mention. This is an example of the so called Ramanujan* summation, where the number $-1/12$ is obtained for the series in the question.

In Cesàro, you define
$$
\sum_{n=1}^\infty a_n=\lim_{N\to\infty}\frac{s_1+\cdots+s_N}N

$$
(if it exists), where $s_k=\sum\limits_{n=1}^k a_n$. Summable in the canonical sense implies Cesàro summable, with the same limit, but not vice versa.

The other canonical summability is the Abel one: you define
$$
\sum_{n=1}^\infty a_n=\lim_{x\to1^-}\sum_{n=1}^\infty a_nx^n.
$$
Again, Cesàro summability implies Abel summability to the same limit, but not vice versa.

To see an example, consider $a_n=(-1)^n$. The usual series doesn't exist. But, using Cesàro or Abel, one can say that $\sum\limits_{n=1}^\infty(-1)^n=\frac12$.