Evaluating the sum $\sum_{i=0}^\infty \frac{F_n}{2015^n}$ 
Find $$\sum_{n=0}^\infty \frac{F_n}{2015^n}$$
Where $F_n$ is the $n$-th Fibonacci number.

Any clues?
 A: Use Binet's formula, $$F_n=\frac 1{\sqrt 5}\left(\left(\frac {1+\sqrt 5}2\right)^n-\left(\frac {1-\sqrt 5}2\right)^n\right)$$  Plug that into your sum and you have two geometric series to sum.
A: There are two geometric series in disguise, since 
$$F_n=\frac{1}{\sqrt{5}}\left(\alpha^n-\beta^n\right),\tag{1}$$ where $\alpha$ and $\beta$ are the real roots of the second degree polynomial $x^2-x-1$. Such closed formula is straightforward to prove by induction on $n$, since both sides of $(1)$ equal $0$ at $n=0$, equal $1$ at $n=1$ and fulfill the recurrence $R_{n+2}=R_{n+1}+R_n$. The same argument leads to the identity:
$$ \forall x:|x|<\frac{\sqrt{5}-1}{2},\qquad \sum_{n\geq 0}F_n x^n = \frac{x}{1-x-x^2},\tag{2} $$
and by evaluating $(2)$ at $x=\frac{1}{2015}$,
$$ \sum_{n\geq 0}\frac{F_n}{2015^n} = \color{red}{\frac{2015}{4058209}}\tag{3}$$
readily follows.
A: Hint: If we write the generating function
$$
f(x) = \sum_{n = 0}^\infty F_n x^n
$$
then because $F_{n+2} = F_{n+1} + F_n$,
$$
f(x) = x + (x + x^2) f(x).
$$
Therefore,
$$
f(x) = \frac{x}{1 - x - x^2}.
$$
Your sum is $f\left(\frac{1}{2015}\right)$.
A: One heavy-handed way of approaching this is to start by proving that
$$
F_n = \frac 1 {\sqrt 5} \left( \left( \frac{1 + \sqrt 5} 2 \right)^n - \left( \frac{1 - \sqrt 5} 2 \right)^n \right).
$$
That can be done by induction on $n$, using the recurrence satisfied by the Fibonacci numbers.
After that, you just need to sum two separate geometric series and add them.
