TITLE: How to prove $\phi=1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_nF_{n+1}}$
QUESTION [6 upvotes]: How do you prove $$\phi=1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_nF_{n+1}}$$ where $\phi$ is the golden ratio and $F_n$ is the $n$th Fibonacci number?
I am aware that $\lim\limits_{n\to\infty}\frac{F_{n+1}}{F_n}=\phi$ and $F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}$ and that this might have something to do with the proof, but I do not know where to start from.

REPLY [2 votes]: If $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$ is given by the continued fraction
$$ \alpha = [a_0;a_1,a_2,\ldots] = a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ldots}}} $$
and 
$$ \frac{p_n}{q_n} = [a_0;a_1,\ldots,a_n] $$
are the convergents of such continued fraction, then $\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}=\pm\frac{1}{q_n q_{n+1}}$ implies
$$ \alpha = a_0+\sum_{n\geq 0}\frac{(-1)^n}{q_n q_{n+1}}.$$
In the particular case $a_0=a_1=a_2=\ldots=1$ the convergents of $\alpha=\frac{1+\sqrt{5}}{2}=1+\tfrac{1}{\alpha}$ are given by ratios of consecutive Fibonacci numbers and the claim is a straightforward consequence of the above Lemma.