Does $\zeta(-1)=-1/12$ or $\zeta(-1) \to -1/12$? I saw NumberPhile channel on Youtube, and they proved $1+2+3+\cdots=-1/12$. Also, I read This.

So, which one is correct
$$\zeta(-1)=-1/12\\ \text{or} \\\zeta(-1) \to -1/12$$
Equivalent to:
$$1+2+3+\cdots=-1/12\\ \text{or} \\1+2+3+\cdots \to -1/12$$

My question: Does it "equal" or "converge"?

Question Explanation:
I mean by "$\to$" "approaches to", like $x\to a $ means $\forall \epsilon>0, |x-a|<\epsilon.$
 A: For the zeta function, it is correct that
$$\zeta(-1)=-\frac{1}{12}$$
The function $\zeta$ is continuous (and continuous in $-1$ in particular).
However, with the usual notion of convergence, we have:
$$1+2+3+4+\cdots = +\infty$$
which is the same as
$$1+2+3+4+\cdots+n \to +\infty \quad\mathrm{as}\quad n\to+\infty$$
To say that
$$1+2+3+4+\cdots = -\frac{1}{12}$$
or
$$1+2+3+4+\cdots+n \to -\frac{1}{12} \quad\mathrm{as}\quad n\to+\infty$$
would require us to specify in what sense this limit is taken. See Wikipedia on $1+2+3+4+\cdots$ for details.
A: $\zeta(z)$ is a meromorphic function with a single pole, which has residue $1$ at $z=1$. $\zeta(-1)=-\frac1{12}$. This fact is used to justify the divergent series
$$
1+2+3+4+\dots=-\frac1{12}
$$
but that divergent series is not why we say that $\zeta(-1)=-\frac1{12}$.

Computation of $\mathbf{\zeta(-1)}$:
Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
$$
\begin{align}
(1-2^{1-x})\zeta(x)\Gamma(x+2)
&=\int_0^\infty\frac{(x+1)xt^{x-1}}{e^t+1}\mathrm{d}t\\
&=\int_0^\infty\frac{(x+1)t^xe^t}{(e^t+1)^2}\mathrm{d}t\\
&=\int_0^\infty\frac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}\mathrm{d}t\tag{1}
\end{align}
$$
Now we can plug in $x=-1$ into $(1)$ to get
$$
\begin{align}
(1-2^2)\zeta(-1)\Gamma(1)
&=\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}\mathrm{d}t\\
&=\int_1^\infty\frac{u-1}{(u+1)^3}\mathrm{d}u\\
&=\int_1^\infty\left(\frac1{(u+1)^2}-\frac2{(u+1)^3}\right)\mathrm{d}u\\
&=\frac14\tag{2}
\end{align}
$$
Since $(1-2^2)\Gamma(1)=-3$, $(2)$ says that
$$
\zeta(-1)=-\frac1{12}\tag{3}
$$
A: It could make sense to say that $\zeta(z)\to-1/12$ as $z\to-1$, but the statement $\zeta(-1)\to-1/12$ as $\text{what}\to\text{what?}$ can make sense only if $\zeta(z)$ is actually equal to $-1/12$.
As for the statement that $1+2+3+\cdots$ bears some relation to $-1/12$, that requires one to define one's "summation method".  But even there, one cannot say $1+2+3+\cdots\to\text{something}$ without at least tacitly saying "$\text{as something}\to\text{something}$, and if the variable that's approaching something does not appear in the expression $1+2+3+\cdots$, then the limit statement would be true only if $1+2+3+\cdots$ is actually equal to the supposed limit.
Later postscript: In view of comments below, may I ask that those who want to comment read what I ACTUALLY wrote before commenting on it?
