Sum of all Fibonacci Numbers equals $-1$?

Question

Context: I was pondering aimlessly when I thought of this, and I don't have any reason to say why this is wrong, other than I know it is wrong

My thoughts go like this: (Before we start, By Fibonacci, I mean the sequence $a_{n}$ where $a_{n} = a_{n-1} + a_{n-2}$, given $a_{0} = a_{1} = 1$)

Let $S$ be the sum of all Fibonacci terms, thus $$S = \sum_{n=0}^{\infty} a_n \tag1$$

So, $$\begin{align} S + S &= a_0 + (a_0 + a_1) + (a_1 + a_2) + (a_2 + a_3) + \ldots \tag2\\[4pt] \implies\quad 2S &= a_0 + a_2 + a_3 + a_4 + a_5 + \ldots \tag3\\[4pt] &= a_0 + S - a_0 - a_1 \tag4\\[4pt] \implies\quad S &= -a_1 = -1 \tag5 \end{align}$$

Problem: Why is this wrong? Although it sounds simple, I brainstormed for an hour but landed on no solid reason. Because of how I subtracted $S$ from $2S$? Or adding $S + S$ because they don't overlap perfectly as I leave a term of the first sum behind? Even so, I don't know why, if the inaccuracy is one of the above two listed, it is wrong.

Help is really, really appreciated

Answer 1

To summarize the discussion in the comments:

You have to be very careful in attaching values to divergent series. Standard arithmetic manipulation leads to "paradoxical" results. For instance, if $S=\sum (-1)^n=1-1+1-1+\cdots$ then plausible arithmetic would tell us that $$S=(1-1)+(1-1)+(1-1)+\cdots=0$$

or $$S=1-(1-1)-(1-1)+\cdots =1 $$

or $$-S=-1+1-1+\cdots =-1+S\implies S=\frac 12$$

and so on.

That said, there are various methods of working with divergent series systematically and these sometimes lead to interesting results.

Analytic continuation is a standard method. If you have a sequence $\{a_n\}$ you can look at the power series $F(z)=\sum a_nz^n$ and try to extend it analytically. If the function has an analytic continuation that covers $z=1$, it might be tempting to define the value of the divergent series to be $F(1)$ or something like that.

For instance, you might note that $$G(z)=\sum z^n=1+z+z^2+\cdots =\frac 1{1-z}$$ when $|z|<1$, but of course that function makes sense for all $z\neq 1$. Taking $z=-1$ gives the series $S$ I mentioned earlier and, following this approach, we might define $S=G(-1)=\frac 12$. And perhaps you feel that makes sense, but $G(2)=-1$ and are you happy with $1+2+2^2+2^3+\cdots =-1$? Of course, letting $T=1+2+2^2+\cdots$, ordinary arithmetic suggests that $2T=T-1$ which, again, implies $T=-1$, so in this sense it is consistent.

In your case, we consider $f_n$, the $n^{th}$ Fibonacci number, and define $$F(z)=\sum_{n=0}^{\infty} f_nz^n=0+z+z^2+2z^3+\cdots$$

As an exercise, you can try to prove that this converges for $z$ in some non-trivial neighborhood of $z=0$ (Hint: use the Binet formula).

The method you discussed makes sense here (at least inside the region of convergence) and we us $f_n=f_{n-1}+f_{n-2}$ to deduce that $$F(z)=0+z+zF(z)+z^2F(z)$$ which implies that $$F(z)=\frac z{1-z-z^2}$$

Note: This differs from what I wrote in the comments in that the numerator is $z$ instead of $1$. That's because this time I started the sequence with $f_0=0$ instead of $f_0=1$ just to bring it in line with more standard notation. This difference was (correctly) pointed out by @ThomasAndrews in the comments and I have changed my function to align with his suggestion.

Now, that function extends analytically to the entire plane, save for the roots of the quadratic denominator.

It's easy to see that $F(1)=-1$ in line with your result.

To stress. None of this should be taken as a sum of the series in any standard sense. The series diverges, it does not converge to any value. But I think the analytic continuation method gives a coherent way to interpret the result.