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 $$ z(x) = 5^{x/2} \sum_{k=0}^{\infty} \binom{-x}{k} \frac{1}{\phi^{x+2k} + (-1)^{k+1} } . $$ 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.

  • $\begingroup$ Cross-posted from math.stackexchange.com/questions/906980/… $\endgroup$
    – Max Muller
    Aug 24, 2014 at 15:14
  • $\begingroup$ A quibble: The way the first example is worded may give a misleading impression, blurring "taking a limit" with "evaluation of meromorphic continuation at a point". If the meromorphic continuation has the desired point at the edge of the circle of convergence, then a non-tangential limit should give the same outcome, yes, but the idea is not restricted to that case... $\endgroup$ Aug 24, 2014 at 16:13
  • $\begingroup$ Duplicate of mathoverflow.net/questions/19410/… $\endgroup$ Aug 24, 2014 at 16:21
  • 2
    $\begingroup$ @DavidSpeyer No, I don't think it's a duplicate. I think we have a different idea of what zeta summation is. You are actually doing summation by means of the analytic continuation of the corresponding dirichlet series. Therefore, you take the limit $ \lim_{x \to 0 } \sum_{n=1}^{\infty} \frac{a_{n}}{n^{x}} $. However, I take the limit $\lim_{x \to -1 } \sum_{n=1}^{\infty} { {a_{n}}^{-x}} $ . The methods are confused quite often with one another, though. $\endgroup$
    – Max Muller
    Aug 24, 2014 at 17:26
  • $\begingroup$ Ah, you are right. Related, but not the same. $\endgroup$ Aug 24, 2014 at 21:13

1 Answer 1


Regarding question number 2, you can write $f_k = \frac{1}{\sqrt{5}} \left( \phi^k - \left(1-\phi\right)^k \right)$, where $\phi^2 = \phi + 1$, $\phi > 0$. Hence $$\sum_{k=1}^{\infty} \frac{f_k}{k^s} = \frac{1}{\sqrt5} \sum_{k=1}^{\infty} \frac{\phi^k - \left(1-\phi\right)^k}{k^s}$$ By the definition of the Polylogarithm, we get $$\sum_{k=1}^{\infty} \frac{f_k}{k^s} = \frac{1}{\sqrt5}\mathrm{Li}_s\left( \phi \right) - \frac{1}{\sqrt5}\mathrm{Li}_s\left( 1-\phi \right)$$

  • $\begingroup$ Thank you. If we do the elementary algebra, we find this amounts to $-1$ as well, when $s=0$. $\endgroup$
    – Max Muller
    Oct 10, 2020 at 17:04

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