# How to prove $\phi=1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_nF_{n+1}}$

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.

• Try to prove $$\sum_{j=1}^{k} \frac{(-1)^{j+1}}{F_{j}F_{j+1}}=\frac{F_{k}}{F_{k+1}}$$ Sep 12 '17 at 13:36
• @David K :I think this series converge to $\frac{1}{\phi}$ Sep 12 '17 at 14:21

Use this formula ($\bf\text{Cassini's identity}$ )$F_n^2 - {F_{n + 1}}{F_{n - 1}} = {( - 1)^{n - 1}}$and we know $(-1)^{n-1}=(-1)^{n+1}$ $$1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_nF_{n+1}}=\\ 1+\sum_{n=1}^{\infty}\frac{F_n^2 - {F_{n + 1}}{F_{n - 1}}}{F_nF_{n+1}}=\\ 1+\sum_{n=1}^{\infty}\frac{F_n^2 }{F_nF_{n+1}}-\frac{ {F_{n + 1}}{F_{n - 1}}}{F_nF_{n+1}}=\\ 1+\sum_{n=1}^{\infty}\left(\frac{F_n }{F_{n+1}}-\frac{ {F_{n - 1}}}{F_n}\right)=\\$$can you go on ?
$$1+\sum_{n=1}^{\infty}\left(\underbrace{\frac{F_n }{F_{n+1}}}_{f_n}-\underbrace{\frac{F_{n-1} }{F_{n}}}_{f_{n-1}}\right)=\\$$
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.