Different type of sum

What we are being asked to evaluate is the sum $S \; = \; \sum_{N=0}^\infty \left(\prod_{m=0}^Na_m\right)^{-1}$ where $a_0=3\,,\,a_1=7\,,\,a_2=47\,,\,a_3=2207$ are the first few terms of the sequence $(a_n)_{n \ge 0}$ defined by the recurrence relation $a_{n+1}^2 \; = \; a_n^2 - 2 \;,\; n \ge 0 \;, \hspace{2cm} a_0 = 3$ It is easy to show by induction that $a_n \; = \; \left(\tfrac{3+\sqrt{5}}{2}\right)^{2^n} + \left(\tfrac{3-\sqrt{5}}{2}\right)^{2^n} \hspace{2cm} n \ge 0$ and further $\sqrt{5}\prod_{m=0}^n a_m \; = \; \left(\tfrac{3+\sqrt{5}}{2}\right)^{2^{n+1}} - \left(\tfrac{3-\sqrt{5}}{2}\right)^{2^{n+1}} \hspace{2cm} n \ge 0$ Thus we deduce that $S \; = \; \sqrt{5}\sum_{n=0}^\infty \left(X^{2^{n+1}} - X^{-2^{n+1}}\right)^{-1} \; = \; \sqrt{5}\sum_{n=1}^\infty \left(X^{2^n} - X^{-2^n}\right)^{-1} \; = \; \sqrt{5}\left[\sum_{n=0}^\infty \left(X^{2^n} - X^{-2^n}\right)^{-1} - \frac{1}{X - X^{-1}}\right]$ where $X = \tfrac{3+\sqrt{5}}{2}$ . Now since $\begin{aligned} \sum_{n=0}^\infty \left(X^{2^n} - X^{-2^n}\right)^{-1} & = \; \sum_{n=0}^\infty X^{-2^n}\big(1 - X^{-2^{n+1}}\big)^{-1} \; = \; \sum_{n=0}^\infty X^{-2^n}\sum_{m=0}^\infty X^{-m 2^{n+1}} \; = \; \sum_{n,m =0}^\infty X^{-(2m+1)2^n} \\ & = \; \sum_{n=1}^\infty X^{-n} \; = \; \frac{X^{-1}}{1 - X^{-1}} \; = \; \frac{1}{X-1} \end{aligned}$ and hence $S \; = \; \sqrt{5}\left[\frac{1}{X-1} - \frac{X}{X^2-1}\right] \; =\; \frac{\sqrt{5}}{X^2-1} \; = \; X^{-1} \; = \; \tfrac12(3 - \sqrt{5})$ making the answer $2 + 3 - 5 = \boxed{0}$ .