Final Round, September 2019

Note that for all $n\ge 0$ the number $a_n$ can be rewritten as follows: $$\begin{array}{rl}{a}_n& =\frac{1}{F_n{F}_{n+2}}=\frac{F_{n+1}}{F_n{F}_{n+2}}\cdot \frac{1}{F_{n+1}}=\frac{F_{n+2}-F_n}{F_n{F}_{n+2}}\cdot \frac{1}{F_{n+1}}\\ & =\left(\frac{1}{F_n}-\frac{1}{F_{n+2}}\right)\cdot \frac{1}{F_{n+1}}=\frac{1}{F_n{F}_{n+1}}-\frac{1}{F_{n+1}{F}_{n+2}}.\end{array}$$ We now get that for each $m\ge 0$ the sum $a_0+a_1+a_2+\cdots +a_m$ equals $$\left(\frac{1}{F_0{F}_1}-\frac{1}{F_1{F}_2}\right)+\left(\frac{1}{F_1{F}_2}-\frac{1}{F_2{F}_3}\right)+\cdots +\left(\frac{1}{F_m{F}_{m+1}}-\frac{1}{F_{m+1}{F}_{m+2}}\right).$$ In this sum all terms cancel, except the first and last. In this way, we get $$ a_0 + a_1 + a_2 + \cdots + a_m = \frac{1}{F_0 F_1} - \frac{1}{F_{m+1} F_{m+2}} = 1 - \frac{1}{F_{m+1} F_{m+2}} < 1.