If only it was (n-1)(n+1)

Note first that $\dfrac{1}{F_{n-1}F_{n+1}} = \dfrac{1}{F_{n-1}F_{n+1}} \times \dfrac{F_{n+1} - F_{n-1}}{F_{n}} = \dfrac{1}{F_{n-1}F_{n}} - \dfrac{1}{F_{n}F_{n+1}}$ .

The sum to $N$ thus telescopes to $\displaystyle\sum_{n=2}^{N} \dfrac{1}{F_{n-1}F_{n+1}} = \dfrac{1}{F_{1}F_{2}} - \dfrac{1}{F_{N}F_{N+1}}$ , which goes to $\boxed{1}$ as $N \to \infty$ .

Here in series numbers are 1,1,2,3,5,8,13..... so,for given equation,for n=1 it would be 0.5 , for n=2 it would be 0.333, for n=3 it woild be 0.1 ,for n=4 it would be 0.0416 ,and for n=5 it would be 0.015 but after it the answer would come in 3 decimal place , not affecting our answer , so adding all the above values we get 0.99 as our answer