Can we use analytic continuation to obtain $\sum_{n=1}^\infty n = b, b\neq -\frac{1}{12}$ 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 $a$.

 A: Let $f_n(z,a)=\frac n{(n+a)^z}$.  Taking the analytic continuation as $z\to0$ on the sum of $f$, we have
$$\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=\frac{a^2}2-\frac1{12}$$
The proof is not so bad.  Notice that
$$\frac\partial{\partial x}\frac1{(xn+a)^{z-1}}=\frac n{(xn+a)^z}$$
And further that
$$\sum_{n=1}^\infty\frac1{(xn+a)^{z-1}}=\frac1{x^{z-1}}\sum_{n=1}^\infty\frac1{(n+\frac ax)^{z-1}}=\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}$$
where we use the Hurwitz zeta function.  Differentiating with respect to $x$ then gives
$$\frac\partial{\partial x}\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}=\frac{-ax^{z-3}\zeta(z,1+\frac ax)-(z-1)x^{z-2}\zeta(z-1,1+\frac ax)}{x^{2z-2}}$$
Let $x=1$ and you end up with
$$\sum_{n=1}^\infty f_n(z,a)=-a\zeta(z,1+a)-(z-1)\zeta(z-1,1+a)$$
And as $z\to0^+$, we get...

$$\lim_{z\to0^+}f_n(z,a)=n\\\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=-a^2\zeta(0)+\zeta(-1)=\frac{a^2}2-\frac1{12}$$

A: Let $f_n(s) = n^{-s}+ (s+1) e^{-(s+1)n} (b-\zeta(-1))$. Then $f_n(-1) = n$ and for $Re(s) > 1$ :  $$F(s) = \sum_{n=1}^\infty f_n(s) = \zeta(s)+(b-\zeta(-1))\frac{s+1}{e^{s+1}-1}$$
It can be continued analytically to the complex plane minus $s=1$ and $s=-1+2ik \pi, k \in \mathbb{Z}^*$ :
$$F(-1) = \zeta(-1) + (b-\zeta(-1))\lim_{s \to -1}\frac{s+1}{e^{s+1}-1}=b$$
