[Largedisplaystyle sum_{n=0}^infty frac{1}{F_{2n+1}+1}=frac{sqrt5}{2}
]
(Largemathbf{Proof:})
Let (phi=dfrac{1+sqrt{5}}{2}) denote the golden ratio. Then consider the partial sum,
[egin{align*} sum_{n=0}^Nfrac{1}{1+F_{2n+1}}&= sum_{n=0}^Nfrac{1}{1+dfrac{phi^{2n+1}+phi^{-(2n+1)}}{sqrt{5}}} \ &= sqrt{5} sum_{n=0}^{N}frac{phi^{2n+1}}{phi^{2(2n+1)}+sqrt{5}phi^{2n+1}+1} \ &=sqrt{5} sum_{n=0}^{N}frac{phi^{2n+1}}{(phi^{2n+1}+phi)left( phi^{2n+1}+dfrac{1}{phi}
ight)}\ &= sqrt{5} sum_{n=0}^{N}frac{phi^{2n+1}}{(phi^{2n}+1)left( phi^{2n+2}+1
ight)} \ &= frac{phisqrt{5}}{1-phi^2}sum_{n=0}^Nleft(frac{phi^{2n}}{1+phi^{2n}}-frac{phi^{2n+2}}{1+phi^{2n+2}}
ight) \ &=sqrt{5}left(frac{phi^{2N+2}}{1+phi^{2N+2}} -frac{1}{2}
ight) end{align*}
]
Let (N oinfty) to get
[Largeoxed{displaystyle sum_{n=0}^inftyfrac{1}{1+F_{2n+1}}=color{blue}{frac{sqrt{5}}{2}}}
]