## Solution

From Cassini’s identity, we have

$$F_{2n+1}^2 - 1 = F_{2n}F_{2n+2}$$

Therefore,

\begin{aligned} \sum_{n=1}^\infty \frac{1}{F_{2n+1} - 1}+ \sum_{n=1}^\infty \frac{1}{F_{2n+1} + 1} &= \sum_{n=1}^\infty \frac{2F_{2n+1}}{F_{2n+1}^2 - 1} \\ &= \sum_{n=1}^\infty \frac{2F_{2n+1}}{F_{2n}F_{2n+2}} \\ &= \sum_{n=1}^\infty \frac{2}{F_{2n}} - \frac{2}{F_{2n+2}}\\ &= \frac{2}{F_2} = 2 \end{aligned}

Using Binet’s formula, we have

\begin{aligned} F_{2n+1} &= \frac{\phi^{2n+1}+\frac{1}{\phi^{2n+1}} }{\sqrt{5}} \\ \end{aligned}

where $\phi = \frac{1 + \sqrt{5}}{2}$.

Using the above, we get

\begin{aligned} \sum_{n=1}^\infty \frac{1}{F_{2n+1} + 1} &= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n+1}}{\phi^{4n+2} + \sqrt{5}\phi^{2n+1} + 1} \\ &= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n+1}}{\phi^{4n+2} + \phi^{2n+2} + \phi^{2n} + 1} \\ &= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n}(\phi^2-1)}{(1+\phi^{2n+2})(1+\phi^{2n})} \\ &= \sum_{n=1}^\infty \sqrt{5} \left( \frac{1}{1+\phi^{2n}} - \frac{1}{1+\phi^{2n+2}} \right) \\ &= \frac{\sqrt{5}}{1 + \phi^2} = \frac{\sqrt{5}}{\sqrt{5}\phi} = \phi - 1 = \frac{\sqrt{5}-1}{2} \end{aligned}