Alternative analytic continuation to zeta, not giving $-\frac{1}{12}$ for sum of integers

Question

Apologies if this has been asked already. Inspired partly by this answer where an $n e^{-\epsilon n}$ rather than $n^s$ regularization was made in the 'evaluation' of $\sum\limits_{n=1}^{\infty}n$ and the number $-\frac{1}{12}$ appeared as the only constant in the answer, and partly by a conversation with a friend claiming that $-\frac{1}{12}$ is the only 'right' way of summing this series, I have the following question:

Does anyone know of or can think of a function $p(s,n)$, $s\in\mathbb{C}$, $n\in\mathbb{N}$ such that $\exists s_0\in\mathbb{C}$ such that $p(s_0,n)=n$, $\forall n\in\mathbb{N}$, where $q(s)=\sum\limits_{n=1}^{\infty}p(s,n)$ is defined and analytic on some domain $D\in\mathbb{C}$ (not containing $s_0$ obviously) and such that $q(s)$ can be analytically continued to $Q(s)$ defined on some larger domain $D_0\supset D$ such that $s_0\in D_0$ and $Q(s_0)\ne-\frac{1}{12}$?

I expect that finding such a function is probably simple, but I have not been successful so far, partly due to not having much familiarity with analytic continuations. (Note that I am not wishing to make any claims or have any arguments about the validity of $1+2+3+4+...\stackrel{?}{=}-\frac{1}{12}$).

Answer 1

$-1/12$ is not really special among the analytic continuation methods, since $$p(s,n) = n^{-s}+(s+1)n^{-2-s} \implies Q(s) = \sum_{n=1}^\infty p(s,n) = \zeta(s)+(s+1) \zeta(s+2) \qquad(Re(s) > 1) $$ $$\implies Q(-1) = \lim_{s \to -1} Q(s)= \zeta(-1)+\lim_{s \to 1}(s-1)\zeta(s)= \frac{11}{12}$$

Thus a natural question is what is special with the $\zeta$ regularization ?

• The $z^n$ regularization is easier to understand : If $$f(z) = \sum_{n=-\infty}^\infty a_n z^n$$ converges for $r < |z| < R$ and can be analytically continued to $z=1$, then $\lim_{z \to 1} f(z) = \lim_{z \to 1} z^k f(z)$ that is $(a_n)$ is the same sequence as $(a_{n-k})$ with respect to power series regularization.

• Now yes the $n^{-s}$ regularization is special in how it acts on shifting the sequence (in a complicated way). But with $$g(s) = \sum_{n=1}^\infty a_n n^{-s}$$ you get that $\lim_{s \to 0} g(s) = \lim_{s \to 0}g(sk)=\lim_{s \to 0} m^{-s}g(s)$ that is $(a_n)$ is the same sequence as $(a_{n^{1/k}})$ and $(a_{n/m})$ for the zeta regularization, and sometimes it is more useful than shifting invariance (multiplicative functions..).

Answer 2

Here's a fun one:

$$\zeta^\star(s,\alpha)=\sum_{n=1}^\infty\frac n{(n+\alpha)^s}$$

Analytically continuing to $s=0$, you'll get

$$\zeta^\star(0,\alpha)=\frac{\alpha^2}2-\frac1{12}$$

And likewise,

$$n=\frac n{(n+\alpha)^0}\forall\alpha$$

This result may be derived by considering another variable:

$$\sum_{n=1}^\infty\frac1{(nx+\alpha)^s}$$

And differentiating w.r.t. $x$. Then factor $x$ out and see the resulting Hurwitz zeta function, which may easily be handled, then letting $x=1$.