Can finite values other than $-1/12$ be assigned to the divergent sum of all natural numbers in a similar way? To define precisely what would be the "similar way", consider the following:
Let $\{a_n\}$ be a sequence such that $\sum_{n=1}^\infty a_n$ is a divergent sum.
Let $f_n(z)$ be complex analytical functions such that:
- There is a $z_0\in\mathbb C$ such that $f_n(z_0) = a_n$, $\forall n\in\mathbb N$.
- There is an open set $S\subset \mathbb C$ and complex analytical function $g(z)$ such that $z_0$ is not its pole and $$\sum_{n=1}^\infty f_n(z) = g(z), ~ \forall z\in S.$$ Obviously, $z_0 \not\in S$.
The value assigned to $\sum_{n=1}^\infty a_n$ is then $g(z_0)$.
In other words, loosely speaking, the procedure can be described as replacing the sequence of numbers $\{a_n\}$ to sequence of functions $\{f_n(z)\}$ such that their sum converges somewhere else, summing them to get another function $g(z)$, analytically extending it to include the point where $f_n(z)=a_n$, evaluating $g(z)$ there and assigning that value to the divergent sum.
Assigning the value $-1/12$ to the divergent sum $\sum_{n=1}^\infty n$ (i.e. $a_n = n$) results from the choice $f_n(z) = n^{-z}$. Then, $z_0 = -1$, $S =\{x+iy|x > 1, y\in\mathbb R\}$, $g(z) = \zeta(z)$ and finally, the assigned value is $g(z_0) = \zeta(-1) = - 1/12$.
Question: Are there other choices of functions $f_n(z)$ that will result in different assigned (finite) values in the above described procedure?
If yes, what are some of the examples and what makes the one resulting in $-1/12$ so special?
If no, how to prove it and can this uniqueness be generalized for any divergent sum to which a finite value can be assigned with the above described procedure?