If the generating function summation and zeta regularized sum of a divergent series exist, do they always coincide? One could assign a value to divergent series by means of several summation methods. One summation method we could consider is the generating function method. Let's sum, for example, the fibonacci series  by means of this method. We consider the generating function of the fibonacci series: $$ g(x) = \sum_{n=1}^{\infty} f_{n} x^{n},  $$ in which the $n$'th Fibonacci number is defined as: $f_{n} = f_{n-1} + f_{n-2} $ and $f_{0} = 0 $ and $f_{1} = 1$. In the following wikipedia article it is explained (amongst many other things) that it is possible to find a closed-form for $g$: $$g(x) = \frac{x}{1-x-x^2} . $$ Therefore, we could state that the divergent series summation by means of the generating function summation method $G$ amounts to taking the following limit: $$ G \Big{(} \sum_{n=1}^{\infty}f_{n} \Big{)} = \lim_{x \to 1} g(x) = -1 .  $$
We could, also try to assign a value to the sum of all Fibonacci numbers by means of zeta function regularisation. A closed form of the function $$ z(x) = \sum_{n=1}^{\infty} f_{n}^{-x} $$ can be found in equation $(5)$ of the following paper by Navas. (Who mistakenly asserts that he is finding the analytic continuation of the Fibonacci Dirichlet series. He is actually doing zeta function regularization of the Fibonacci series. The two methods are confused quite often, though.) He finds that $$ z(x) = 5^{x/2} \sum_{k=1}^{\infty} \binom{-x}{k} \frac{1}{\phi^{x+2k} + (-1)^{k+1} } . $$ The zeta regularized sum $R$ of the Fibonacci series is therefore $$ R \Big{(} \sum_{n=1}^{\infty} f_{n} \Big{)} = \lim_{x \to -1} z(x) = \frac{1}{\sqrt{5}} \Big{(} \frac{1}{\phi^{-1} -1} + \frac{1}{\phi + 1} \Big{)} = -1 .$$ We have thus found that the generating function summation and the zeta regularized sum of the Fibonacci series coincide (define $F := \sum_{n=1}^{\infty} f_{n} )$ : $$ G(F) = -1 = R(F) .$$
This does not always happen, though. If we define the sum of natural numbers $ N = \sum_{n =1}^{\infty} n $, then $G (N) $ does not exist. This is because the corresponding generating function amounts to $$ p(x) = \sum_{n=1}^{\infty} n x^{n} = \frac{1}{ (1-x)^{2} } , $$ for which $\lim_{x \to 1 } p(x) $ does not exist. The zeta regularized sum does exist, however. We have the notorious equation $R(N) = - \frac{1}{12} $ (see this page).
Question (1) is now: Do the $G$ and $R$ summation methods of a divergent series always coincide, if both methods lead to a finite number that can be assigned to the divergent series at hand?
Question (2): Is there a closed form of the actual analytic continuation of the Fibonacci dirichlet series $ d(x) = \sum_{n=1}^{\infty} \frac{ f_{n} }{ n^{x} } $ ?
Bonus question: are there any references for collections of generating function expansion, zeta function analytic continuations and analytic continuations of dirichlet series? For the first group of functions there is the book Generatingfunctionology by Wilf, but I can't find any big overview papers/books/articles on the last two groups of functions.  
 A: Just an idea.
The zeta-function has a laurent-series-representation as a sum of the term $$\zeta_a(x)=-{1\over1-x}$$ and of the series $$\zeta_b(x) = 1/2 + 0.081x-0.0031x^2+... = \sum_{k=0}^\infty c_k x^k$$ such that
$$\zeta (x)=\zeta_a(x)+\zeta_b(x)$$
If I recall correctly, the series $\zeta_b(x)$ is entire - this means, that in the case of arguments $x$ where the zeta becomes divergent, the divergent part is completely eaten by the $\zeta_a(x)$ part - and the behave of the geometric series is completely accepted as regular even in the divergent cases by the expression as fraction ${1\over1-x}$ with the sole exception of the case where $x=1$.          
This is not a proof for the identity of the zeta-regularization and the power-series/generating function, but it shows at least a strong formal relation ship: the zeta -even in the divergent cases as formal sum of two powerseries regularly computable- and the power series of the generating function.
I think someone with more formal background might be able to put this two things together.
