Does $\zeta(-1)=-1/12$ or $\zeta(-1) \to -1/12$? [duplicate]
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 $\forall \epsilon>0, |x-a|<\epsilon.$
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.
$\zeta(z)$ 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 $$ \zeta(-1)=-\frac1{12}\tag{3} $$