TITLE: How to prove the Fibonacci sum $\sum \limits_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$
QUESTION [29 upvotes]: We are familiar with the nifty fact that given the Fibonacci series $F_n = 0, 1, 1, 2, 3, 5, 8,\dots$ then $0.0112358\dots\approx 1/89$. In fact,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
How do we prove that, more generally, for $p > 1$ then,
$$\sum_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$$
(The above simply was the case $p = 10$.)

REPLY [17 votes]: Let $f(z) = \sum\limits_{n=0}^\infty F_nz^n$
Then 
$$\begin{align}zf(z) + z^2f(z) &= \sum_{n=1}^\infty {F_{n-1}z^n} + \sum_{n=2}^\infty F_{n-2}z^n\\
&=F_0z + \sum_{n=2}^\infty (F_{n-2}+F_{n-1})z^n 
\\&=F_0z + (f(z)-F_0 - F_1z)
\end{align}$$
Using $F_0=0, F_1=1$, we can solve for $f(z)$:
$$f(z)=\frac{z}{1-z-z^2}$$
Now substitute $z=\dfrac{1}p$.
[Note, this doesn't prove that the series converges, only that, if the series converges, it converges to this value.  It might require a little more effort to prove that it converges for $z=\dfrac{1}{p}$.]