Unsure about result involving Fibonacci Numbers $\sum_{n = 0}^\infty F_n$

Question

I just starting working through a text on Moment Generating Functions. One elementary example is that of the Fibonacci Sequence and determining it's Moment Generating Function. I've found a result for the sum of all Fibonacci Numbers that I'm unsure of: $$ \sum_{n = 0}^\infty F_n = - 1 $$ Okay, so some basics to begin with:

Here, let $F_n$ be the n-th Fibonacci Number. The Fibonacci Sequence is defined via the recursion: $$ F_{n + 2} = F_{n + 1} + F_n $$ Where $F_0 = 0, F_1 = 1$ and $n \geq 0$

The method taken is to define the Moment Generating Function $F(x)$ as $$ F(x) = \sum_{n = 0}^\infty F_nx^n $$ Where $x \in \mathbb{R}$. When you apply this to the recursive relationship, you find the moment generating function to be $$ F(x) = \frac{x}{1 - x - x^2} $$ And so, $$ \frac{x}{1 - x - x^2} = \sum_{n = 0}^\infty F_n x^n $$ And so, evaluating at $x = 1$, we find $$ \frac{1}{1 - 1 - 1^2} = \sum_{n = 0}^\infty F_n 1^n \longrightarrow \sum_{n = 0}^\infty F_n = -1 $$ Which seems like an odd result given $F_n \geq 0$. With only being an amateur mathematician, I thought this may be similar to the result: $$ \sum_{n = 1}^\infty n = -\frac{1}{12} $$ Regardless, my first thought is that the result I found could be the result of evaluating $F(x)$ at a value of $x$ in which it's non-convergent. I decided to apply the Ratio Test: $$ r = \lim_{n \rightarrow \infty} \left| \frac{F_{n + 1}x^{n + 1}}{F_{n }x^{n}} \right| = \lim_{n \rightarrow \infty} \left| \frac{F_{n + 1}}{F_{n }} x\right| = \psi\left|x\right| $$ And so for convergence we require $r < 1$ or $$ \psi\left|x\right| < 1 \longrightarrow \left|x\right| < \frac{1}{\psi} \approx 0.618 $$ And so, unless I'm mistaken $F(x)$ is only convergent for $\left|x\right|< \frac{1}{\psi}$ and thus at $x = 1$ (as I applied) is invalid.

Is this reasoning correct? and if so, is there any validity to the result of $-1$?

Thanks!

Answer 1

Yes. The series is not convergent so you cannot put $x=1$ in the series. You must have $$\sum_{n=1}^{\infty}F_n = \infty.$$ However, it is the value of the analytic continuation of the function $$f(z)=\sum_{n=1}^{\infty}F_nz^n$$ initially defined on the ball $B_{1/\psi}(0)$.

Answer 2

The series is divergent. However, what you are finding is the Borel sum of the series. As the Wikipedia article noted: it is a method for summing divergent asymptotic series, and in some sense gives the best possible sum for such series

Note that

\begin{align*} \sum_{n=0}^{\infty}F_{n} = & \sum_{n=0}^{\infty}\frac{F_{n}}{n!}n!\\ =& \sum_{n=0}^{\infty}\frac{F_{n}}{n!}\int_{0}^{\infty} t^{n}e^{-t}dt \\ \stackrel{\mathcal{B}}{=}&\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{F_{n}}{n!} t^{n}e^{-t}dt \tag{*}\\ =& \frac{1}{\sqrt{5}}\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{(\phi^{n}-(-\phi)^{-n})}{n!} t^{n}e^{-t}dt \quad \textrm{ (Binet's formula)} \\ =&\frac{1}{\sqrt{5}}\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{((t\phi)^{n}-(-t\phi^{-1})^{n})}{n!} e^{-t}dt\\ =& \frac{1}{\sqrt{5}}\int_{0}^{\infty}(e^{t\phi}-e^{-t\phi^{-1}}) e^{-t}dt\\ =& \frac{1}{\sqrt{5}}\left[\int_{0}^{\infty}e^{-t(1-\phi)}dt-\int_{0}^{\infty}e^{-t(1+\phi^{-1}})dt\right]\\ =& \frac{1}{\sqrt{5}}\left[ \frac{1}{1-\phi} - \frac{1}{1+\phi^{-1}}\right]\\ =& -1 \end{align*}

Hence

$$ \sum_{n=0}^{\infty}F_{n} = -1 \left(\mathcal{B}\right)$$

The notation $\left(\mathcal{B}\right)$ means that the Borel sum of $\sum_{n=0}^{\infty}F_{n} $ converges at $-1$

The "misstep" in (*) is that we are allowing the interchange of integral and series despite the divergence of the series but this is perfectly valid in Borel sums.

Answer 3

$$S_p=\sum_{n = 0}^p F_n =F_{p+2}-1 $$ and its generating function is $$\frac{x^3}{(1-x) \left(1-x-x^2\right)}$$ given by Simon Plouffe.

Expanding around $x=1$ as you wanted to do $$\frac{x^3}{(1-x) \left(1-x-x^2\right)}=\frac{1}{x-1}+2 (x-1)-5 (x-1)^2+O\left((x-1)^3\right)$$

Have a look at sequence $A000071$ in $OEIS$