# (Non-)Canonicity of using zeta function to assign values to divergent series [duplicate]

$$1 + 2 + 3 + \cdots = -1/12,$$

and I wanted to convince myself there was nothing particularly unique about this identity or the Riemann zeta construction. More precisely, this identity only really makes sense if you think of an integer $n$ as being the specialization at $z=-1$ of the function $n^{-z}$. So here's a question:

For any complex number $c$, does there exist a domain $\Omega \subset \mathbb{C}$, and analytic functions $F(n, s)_{n\in\mathbb{N}}$ and $f(s)$ on $\Omega$, such that the following hold

i. $F(n,0) = n$

ii. $\sum_{n=1}^\infty F(n,s) = f(s)$ on $\Omega$ in some reasonable sense (maybe converges uniformly on compact subsets of $\Omega$?)

iii. $f$ can be extended holomorphically to some domain containing both $\Omega$ and $0$ such that $f(0) = c$.

So shifting the Euler series and Riemann zeta would be such a construction for $c=-1/12$. As the question stands, I feel that the answer is almost certainly yes, although to be fair the functions $n^{-s}$ have a lot more structure than "holomorphic functions on some domain". So a follow-up question would be: are there "natural" additional constraints for which the answer to this question is No?

I apologize that this is kind of open-ended, but the goal is to convince myself that there is nothing particularly canonical about $-1/12$ (or to hear an explanation of why it is canonical).

Note that $1+2+3+\ldots = (1+1+1+\ldots)+(0+1+2+3+\ldots)$; so if you can regularize $1+1+1+\ldots$ into something nonzero, then you can shift your result away from $-1/12$. Specifically, try $$1+2+3+\ldots=\sum_{n=1}^{\infty}\left(n^{z} + (n-1)^{z+1}\right)\big\vert_{z=0}=\zeta(-z)+\zeta(-z-1)\big\vert_{z=0}=\zeta(0)+\zeta(-1)=-\frac{7}{12},$$ where the sum converges for $\Re (z) < -2$.
• Is it really valid to split the infinite sum into two separate sums, when all three sums diverge at the point of interest ($z=0$)? Jun 27 '19 at 21:09
• "Valid" in what sense? The divergent infinite sum $1+2+3+\ldots$ is meaningless; at best it represents a divergent sequence of partial sums. The infinite sum $\sum_{n=1}^{\infty} (n^z + (n-1)^{z+1})$ is a function of $z$ defined for some values of $z$ (where it's absolutely convergent), and with the same (divergent) sequence of partial sums as the first expression at $z=0$. Regularization just means replacing a meaningless sum with one that's at least meaningful somewhere. But that's never "valid"... not least because you can do it many different ways and arrive at different answers. Jul 10 '19 at 2:26
• I think there might be an error. You said it converges for $\mathfrak{R}(z) < -2$, but $z=-3$, for example, yields $(1-1)^{-3+1}=0^{-2}$. Jul 10 '19 at 3:34