Fibonacci series

Relevant wiki: Fibonacci Sequence

If we write $\phi = \tfrac12(\sqrt{5}+1)$ , then (by BInet's Formula) $F_{2n+1} + 1 \; = \; \frac{1}{\sqrt{5}}\big[\phi^{2n+1} + \sqrt{5} + \phi^{-2n-1}\big] \; = \; \frac{1}{\sqrt{5}}(\phi^n + \phi^{-n})(\phi^{n+1} + \phi^{-n-1})$ and hence $\begin{array}{rcl} \displaystyle \frac{1}{F_{2n+1}+1} & = & \displaystyle \frac{\sqrt{5}}{(\phi^n + \phi^{-n})(\phi^{n+1} + \phi^{-n-1})} \; = \; \frac{\sqrt{5} \phi^{2n+1}}{(\phi^{2n}+ 1)(\phi^{2n+2} + 1)} \\ & = & \displaystyle \sqrt{5}\left(\frac{1}{\phi^{2n}+1} - \frac{1}{\phi^{2n+2}+1}\right) \end{array}$ from which it is clear that $\sum_{n=0}^N \frac{1}{F_{2n+1}+1} \; = \; \sqrt{5}\left(\frac12 - \frac{1}{\phi^{2N+2}+1}\right)$ and hence the infinite sum is $\sum_{n=0}^\infty \frac{1}{F_{2n+1}+1} \; = \; \frac{\sqrt{5}}{2}$ making the answer $5+2=\boxed{7}$ .