A Fibonacci series

Using the definition of the Fibonacci sequence we see that: $F_n=F_{n+1}-F_{n-1}$ . We can use this information to turn the series of the problem into a telescoping series.

$\sum_{n=2}^\infty \frac{F_n}{F_{n+1} \cdot F_{n-1}}=\sum_{n=2}^\infty \frac{F_{n+1}-F_{n-1}}{F_{n+1} \cdot F_{n-1}}=\sum_{n=2}^\infty \left (\frac{1}{F_{n-1}} - \frac{1}{F_{n+1}}\right )=\lim_{N \to \infty} \sum_{n=2}^N \left (\frac{1}{F_{n-1}}-\frac{1}{F_{n+1}} \right )=$

$=\lim_{N \to \infty} \left (\frac{1}{F_1}-\frac{1}{F_3}+\frac{1}{F_2}-\frac{1}{F_4}+\frac{1}{F_3}-\frac{1}{F_5}+...+\frac{1}{F_{N-2}}-\frac{1}{F_N}+\frac{1}{F_{N-1}}-\frac{1}{F_{N+1}}\right )=\lim_{N \to \infty} \left (\frac{1}{F_1}+\frac{1}{F_2}-\frac{1}{F_N}-\frac{1}{F_{N+1}}\right )=\frac{1}{F_1}+\frac{1}{F_2}-0-0=1+1=2$

Similar solution with @Santi Spadaro 's

$\begin{aligned} S & = \sum_{n=2}^\infty \frac {F_n}{{\color{#3D99F6}F_{n+1}}F_{n-1}} & \small \color{#3D99F6} \text{Note that }F_k = F_{k-1} + F_{k-2} \\ & = \sum_{n=2}^\infty \frac {F_n}{{\color{#3D99F6}\left(F_n+F_{n-1}\right)}F_{n-1}} \\ & = \sum_{n=2}^\infty \left(\frac 1{F_{n-1}} - \frac 1{F_n+F_{n-1}} \right) \\ & = \sum_{\color{#3D99F6}n=2}^\infty \left(\frac 1{F_{\color{#3D99F6}n-1}} - \frac 1{F_{\color{#3D99F6}n+1}} \right) \\ & = \sum_{\color{#D61F06}n=1}^\infty \left(\frac 1{F_{\color{#D61F06}n}} - \frac 1{F_{\color{#D61F06}n+2}} \right) \\ & = \frac 1{F_1} - \frac 1{F_3} + \frac 1{F_2} - \frac 1{F_4} + \frac 1{F_3} - \frac 1{F_5} + \frac 1{F_4} - \frac 1{F_6} + \cdots \\ & = \frac 1{F_1} - {\color{#3D99F6}\cancel{\frac 1{F_3}}} + \frac 1{F_2} - {\color{#D61F06}\cancel{\frac 1{F_4}}} + {\color{#3D99F6}\cancel{\frac 1{F_3}}} - \cancel{\frac 1{F_5}} + {\color{#D61F06}\cancel{\frac 1{F_4}}} - \cancel{\frac 1{F_6}} + \cdots \\ & = \frac 1{F_1} + \frac 1{F_2} = \boxed 2 \end{aligned}$