<s>[INST] What is the value of $\sum_{n=1}^\infty \frac{1}{F_n F_{n+4}}$ where $F_n$ is nth Fibonacci number? I found this problem from my old note, with the memo 'The answer is very interesting."
But I forgot how to solve it! 
I rummaged through all my bookshelf, but I couldn't find any clue.
The only thing I remember is 

*

*Partial fraction decomposition is used.

*It would be easier to solve by using $\sum_{n=1}^\infty \frac{1}{F_n F_{n+2}}=\sum_{n=1}^\infty \left( \frac{1}{F_n F_{n+1}} - \frac{1}{F_{n+1} F_{n+2}} \right) = 1$.

*The solution is not that complicated.

Would you help me?
 [/INST] It can be reduced to $\sum \frac1{F_n F_{n+2}}$ as follows.
Multiply the numerator and denominator by $F_{n+2}$, and write this in terms of $F_n$ and $F_{n+4}$ by using that the sequences $(F_{2n})$ and $(F_{2n+1})$ are linear recurrences.
You will obtain something like
$$\frac1{F_{n}F_{n+4}} = \frac{F_{n+2}}{F_{n}F_{n+2}F_{n+4}} = \frac{a}{F_nF_{n+2}} + \frac b{F_{n+2} F_{n+4}}$$
so that we obtain roughly $(a+b) \sum \frac1{F_n F_{n+2}}$.

Details:
In fact, $F_{n+4} = c_1 F_{n+2} + c_2 F_n$ with $c_1$ and $c_2$ such that the roots of the polynomial $x^2 - c_1 x - c_2$ are $\varphi^2$ and $\overline{\varphi}^2 = \varphi^{-2}$. In particular:

*

*$c_1 = \varphi^2 + \varphi^{-2} = (\varphi - \varphi^{-1})^2 - 2 =  (\sqrt 5 F_1)^2 - 2 = 3$;

*$c_2 = -1$.

We obtain
$$\begin{align*}
\sum_{n=1}^\infty \frac1{F_n F_{n+4}}
&= \sum_{n=1}^\infty \frac{F_{n+2}}{F_n F_{n+2} F_{n+4}} \\
&= \sum_{n=1}^\infty \frac{\frac1{c_1} F_{n+4} - \frac{c_2}{c_1} F_{n+1}}{F_n F_{n+2} F_{n+4}} \\
&= \frac13 \sum_{n=1}^\infty \left(\frac{1}{F_n F_{n+2}} + \frac{1}{F_{n+2} F_{n+4}} \right) \\
&= \frac23 \sum_{n=1}^\infty \frac{1}{F_n F_{n+2}} - \frac13 \frac1{F_1 F_3} - \frac13 \frac1{F_2 F_4} \\
&= \frac23 - \frac16 - \frac19 \\
&= \frac7{18} \,.
\end{align*}$$
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