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I tried to follow generating function for Fibonacci numbers proof, which proves this:

$$\sum_{n=0}^{\infty} F_{n+2} z^n = \frac{1}{1-(z+z^2)}. $$

Everything seems to be clear, but I still have a question:

Does this mean that I can calculate an infinit sum of Fibonacci numbers multiplied by $z^n$?

Say if I set $z=1$ does this mean that $$\sum_{n=0}^{\infty} F_{n+2} 1^n = \frac{1}{1-(1+1^2)}= \frac{1}{-2}=-0.5. $$

Somehow it doesn't feel right. Could you point me to the mistake I did?

Edit

Apparently there is a constraint on $z$ which is missing. Taking as an example one of the proofs, from which operation does this constraint should come? $$\sum_{n=0}^{\infty} F_{n+2} z^n = \sum_{n=0}^{\infty} F_{n+1} z^n + \sum_{n=0}^{\infty} F_{n} z^n $$

$$\implies \sum_{n=2}^{\infty} F_{n} z^{n-2} = \sum_{n=1}^{\infty} F_{n} z^{n-1} + g(z) $$

$$\implies \frac{1}{z^2}\sum_{n=2}^{\infty} F_{n} z^{n} =\frac{1}{z} \sum_{n=1}^{\infty} F_{n} z^{n} + g(z) $$

$$ \implies \frac{1}{z^2}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z^2}-\frac{F_1}{z}= \frac{1}{z}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z} + g(z) $$

$$ \implies \frac{g(z)}{z^2}-\frac{1}{z^2}-\frac{1}{z} = \frac{1}{z}g(z)-\frac{1}{z} + g(z) $$

$$ \implies g(z) = \frac{1}{1-(z+z^2)}. $$

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Alexandr
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    Generating functions are formal power series, so they do not necessarily converge. (You get the same sort of problem if you say $\sum_{k=0}^\infty x^k=\frac{1}{1-x}$ and plug in $x=2$... you get $\sum_{k=0}^\infty 2^k=-1$. This is because the power series does not converge at $x=2$.) – kccu Nov 10 '16 at 15:41
  • The constraint comes from the very first equation – unless $z$ satisfies the constraint, the three series don't converge, so putting them in an equation is nonsense. But the exact form of the constraint doesn't become clear until the very last step, when you have to pick $z$ so the denominator doesn't vanish. – Gerry Myerson Nov 11 '16 at 08:34

1 Answers1

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Power series (ie. series of the form $\sum_n a_n z^n$) have a radius of convergence, ie. a radius $r$ such as $|z| < r \Rightarrow $ the series converges.

It only makes sense to apply the series to $z$ (strictly) within the radius of convergence. In your case the radius of convergence is $\phi^{-1}$ (the inverse of the golden ratio), and it makes no sense evaluating the series with $1$, as there is no convergence.

However, the right hand side is well defined for $z=1$ (actually everywhere except 2 poles at $\phi$ and $\overline{\phi}$), this is a coincidence which essentially means that the function defined by the series can be analytically prolonged to the complex plane, minus these two poles.

However the series itself does not exhibit convergence at $1$, it can only be continued at that point.

Ulysse Mizrahi
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  • Ok, so the formula must be $\sum_{n=0}^{\infty} F_{n+2} z^n = \frac{1}{1-(z+z^2)}, |z| < 1. $. But in which place of the reasoning (which I added to the question) this constraint is missed? – Alexandr Nov 11 '16 at 08:24
  • $|z|<1$ isn't good enough. Do you know, Alexandr, how fast the Fibonacci numbers grow? – Gerry Myerson Nov 11 '16 at 08:32
  • @GerryMyerson, yes I know they grow fast, but I try to understand and interpret the reasoning, which seems to be correct. So which constraint is good enough, then? – Alexandr Nov 11 '16 at 08:39
  • If you know how fast they grow, then the ratio test will tell you the appropriate constraint. – Gerry Myerson Nov 11 '16 at 08:45
  • So, again: do you know how fast the Fibonacci numbers grow? – Gerry Myerson Nov 13 '16 at 05:38
  • Had a brain fart on the radius of convergence, edited it out and removed the part about Abel convergence. – Ulysse Mizrahi Nov 14 '16 at 09:13