An infinite sum consisting of Fibonacci numbers:

\[\displaystyle\sum_{n\geq0}\binom{2n}{n}\dfrac{F_n}{8^n}=\sqrt{\dfrac{2}{5}}\]

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For any integer \(n\geq3\),

\[\left\lfloor\dfrac{1}{1-\displaystyle\prod_{k=n}^\infty\left(1-\dfrac{1}{F_k}\right)}\right\rfloor=F_{n-2}\]
where \(F_n\) is the \(n\)-th Fibonacci number.
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