$a_0=\frac{1}{2},a_1=\frac{1}{3}$ $a_0=1-\frac{1}{F_1F_2},a_0+a_1=1-\frac{1}{F_2F_3}$ We will prove, that $a_0+...+a_m=1-\frac{1}{F_{m+1}F_{m+2}}$ Let it is true for $m-1$: $a_0+...a_{m-1}= 1- \frac{1}{F_mF_{m+1}}$ $a_0+...a_{m-1}+a_m=1- \frac{1}{F_mF_{m+1}}+\frac{1}{F_mF_{m+2}}=1- \frac{F_{m+2}-F_{m+1}}{F_mF_{m+1}F_{m+2}}=1-\frac{F_m}{F_mF_{m+1}F_{m+2}}=1-\frac{1}{F_{m+1}F_{m+2}}$ So $a_0+...a_{m-1}+a_m<1$

Note that \[\frac{1}{F_{m}F_{m+2}}=\frac{F_{m+2}-F_{m}}{F_{m+1}(F_{m}F_{m+2})}=\frac{1}{F_{m+1}}\left(\frac{1}{F_{m}}-\frac{1}{F_{m+2}}\right)\]So, we have \[\sum_{i=0}^m \frac{1}{F_iF_{i+2}}=\sum_{i=0}^m \frac{1}{F_{i+1}}\left(\frac{1}{F_{i}}-\frac{1}{F_{i+2}}\right)=\frac{1}{F_0F_1}-\frac{1}{F_{m+1}F_{m+2}}=1-\frac{1}{F_{m+1}F_{m+2}}<1 \]since the summation is a telescoping series. Therefore $\forall m\ge 0$, we see that it approaches $1$ but it will never exceeds $1$ since $1$ is still subtracted by a small term $\frac{1}{F_{m+1}F_{m+2}}$. $\quad \blacksquare$