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.

Here's a reasonably efficient low-tech approach (i.e., not using Binet's formula or generating functions).
Prove by induction that
$$ \sum_{k=0}^n \frac{F_k}{2^k} = 2 - \frac{F_{n+3}}{2^n} \tag{1} $$
and that
$$ nF_n \le 2^n \tag{2} $$
(For (2) you might find it convenient to assume $n\ge 6$.)  From (2) it follows that $\frac{F_{n+3}}{2^n} \le \frac8{n+3} \to 0$, so the partial sums in (1) converge.
