Infinite sum of reciprocal shifted Fibonacci numbers

I found on Wikipedia the following infinite sum :

$$\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}=\frac{\sqrt{5}}{2}$$

There is no reference for this sum in the article and I couldn't find it anywhere else .I have no idea how to prove it so I'm asking for help .

Thank you for your time to help me !

Solution 1:

We can use Binet's formula for $F_{2k+1}$:- $$F_{2k+1}=\frac{\phi^{2k+1}-(-\phi)^{-(2k+1)}}{\sqrt{5}}$$ where $\phi=\frac{1+\sqrt{5}}{2}$.

The summation (after substitution of $F_{2k+1}$ and partial fraction decomposition) becomes a telescoping sum $$\begin{align}\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}&=\sum_{k=0}^{\infty} \frac{\sqrt{5}\phi^{2k+1}}{\sqrt{5}\phi^{2k+1}+\phi^{4k+2}+1}\\&=\sum_{k=0}^{\infty} \left(\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{2k+1}+\sqrt{5}+1}-\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{2k+1}+\sqrt{5}-1}\right)\\&=\frac{\sqrt{5}}{2}-x_1+x_1-x_2+x_2+\cdots+\lim_{k\rightarrow\infty} x_k\\&=\frac{\sqrt{5}}{2}\end{align}$$ where $$x_1=\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{1}+\sqrt{5}-1}=\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{3}+\sqrt{5}+1},x_2=\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{3}+\sqrt{5}-1}=\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{5}+\sqrt{5}+1},\cdots$$ and $\lim_{k\rightarrow\infty} x_k=0$.