TITLE: Prove that $\sum\limits_{n=0}^{\infty}\frac{F_{n}}{2^{n}}= \sum\limits_{n=0}^{\infty}\frac{1}{2^{n}}$
QUESTION [9 upvotes]: I came up with this identity in high school, and I can't remember how I proved it :P  Does anyone know how I would go about doing this?
$$\sum_{n=0}^{\infty}\frac{F_{n}}{2^{n}}= \sum_{n=0}^{\infty}\frac{1}{2^{n}}$$

REPLY [14 votes]: Just for fun, here is a proof using probability. 
Let $N$ be the number of times you 
toss a fair coin until you get two heads in a row.
Here are some outcomes for small $N$ values:

$N=2\quad$ HH 
$N=3\quad$ THH
$N=4\quad$ TTHH, HTHH
$N=5\quad$ TTTHH, THTHH, HTTHH
$N=6\quad$    TTTTHH, TTHTHH, THTTHH, HTTTHH, HTHTHH
etc.

The outcomes of length $n$ are formed in two ways: by sticking a T
in front of an outcome of length $n-1$ or sticking HT in front of 
an outcome of length $n-2$. Thus,  the number of outcomes
of length $n$ is $F_{n-1}$.
Since the probabilities must add to one, we have 
$$\sum_{n=2}^\infty {F_{n-1}\over 2^n}=1$$
which is equivalent to the required identity.