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