<s>[INST] Find the sum of an infinite series of Fibonacci numbers divided by doubling numbers. How would I find the sum of an infinite number of fractions, where there are Fibonacci numbers as the numerators (increasing by one term each time) and numbers (starting at one) which double each time as the denominators?
I'm assuming this has something to do with limits.
 [/INST] Using Moivre-Binet
$$
F_n = \frac{\phi^n - \psi^n}{\phi - \psi}
$$
with $\phi$ and $\psi$ being the positive and negative root of the golden ratio equation $x^2 - x - 1 = 0$,
we get
\begin{align}
\sum_{n=0}^\infty \frac{F_n}{2^n}
&= 
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \frac{\phi^n-\psi^n}{2^n} \\
&= 
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \left(\frac{\phi}{2}\right)^n -
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \left(\frac{\psi}{2}\right)^n  \\
&= 
\frac{1}{\phi - \psi} \frac{1}{1 - \frac{\phi}{2} } - 
\frac{1}{\phi - \psi} \frac{1}{1 - \frac{\psi}{2} } \\
&= 
\frac{2}{\phi - \psi} \left( 
\frac{1}{2 - \phi} - \frac{1}{2 - \psi} 
\right) \\
&= 
\frac{2}{\phi - \psi} \left( 
\frac{(2-\psi) - (2-\phi)}{(2 - \phi)(2-\psi)}
\right) \\
&= 
\frac{2}{\phi - \psi} \left( 
\frac{\phi - \psi}{4 - 2 \phi - 2 \psi + \phi \psi}
\right) \\
&= 
\frac{2}{4 - 2 (\phi + \psi) + \phi \psi} \\
&= 
\frac{2}{4 - 2 + (-1)} \\
&= 2
\end{align}
Note: The geometric series converge because $\lvert \phi/2 \rvert < 1$ and $\lvert \psi/2 \rvert < 1$. 
Appendix:
$$
x^2 - x - 1 = 0 \iff x = \frac{1 \pm \sqrt{5}}{2}
$$
Thus 
$$
\phi + \psi = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1
$$
and
$$
\phi \psi = \frac{1 + \sqrt{5}}{2} \frac{1 - \sqrt{5}}{2} = \frac{1 - 5}{4} = -1
$$
</s>