Evaluating the sum $\sum_{i=0}^\infty \frac{F_n}{2015^n}$

Question

Find $$\sum_{n=0}^\infty \frac{F_n}{2015^n}$$

Where $F_n$ is the $n$-th Fibonacci number.

Any clues?

Answer 1

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 $f\left(\frac{1}{2015}\right)$.

Answer 2

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.

Answer 3

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.

Answer 4

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.