complex analysis - Can we use analytic continuation to obtain $sum_{n=1}^infty n = b, bneq -frac{1}{12}$

Intuitive question

It is a popular math fact that the sum definition of the Riemann zeta function:
$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$
can be extended to the whole complex plane (except one) to obtain $\zeta(-1)=-\frac{1}{12}$. The right hand side for the above equation in $-1$ becomes the sum of the natural numbers so in some sense we have obtained a value for it. My question is: is this value depending on the choice of the Riemann zeta function as the function to be analytically continued, or do we always get $-\frac{1}{12}$?

Proper formulation

let $(f_n:D\subset \mathbb{C}\rightarrow \mathbb{C})_{n\in \mathbb{N}}$ be a sequence of functions and $a \in \mathbb{C}$ with $\forall n\in \mathbb{N}: f_n(a) = n$ and
$$f(z):=\sum_{n=0}^\infty f_n(z)$$
convergent on a part of the complex plane, such that it can be analytically continued to a part of the plane that contains $a$. Does it then follow that, under this continuation, $f(a)=-\frac{1}{12}$ and why (or can you give a counterexample)?

Examples

• The case of the Riemann zeta function is the case where $\forall n
  \in \mathbb{N}: f_n(s) = \frac{1}{n^s}$ and $a=-1$


• The case where $\forall n \in \mathbb{N}: f_n(z) = \frac{n}{z^n}$ and $a=1$ does yield the sum of all natural numbers but it's continuation $\frac{z}{(z-1)^2}$ has a pole at $a$.