TITLE: Sum of inverse of Fibonacci numbers
QUESTION [9 upvotes]: If $F(n)$ is the nth Fibonacci number, How can I prove that:
$$\sum_{i=1}^{\infty} \frac{1}{F(i)}\approx 3.36\, .$$

REPLY [7 votes]: $$
\begin{align}
F_n
&=\frac{\phi^n-(-1/\phi)^n}{\sqrt5}\\
&=\frac{\phi^n}{\sqrt5}\left(1-\left(-\frac1{\phi^2}\right)^n\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{F_n}
&=\sum_{n=1}^\infty\frac{\sqrt5}{\phi^n}\left(1+\left(-\frac1{\phi^2}\right)^n+\left(-\frac1{\phi^2}\right)^{2n}+\left(-\frac1{\phi^2}\right)^{3n}+\cdots\right)\\
&=\sqrt5\left(\frac1{\phi-1}-\frac1{\phi^3+1}+\frac1{\phi^5-1}-\cdots\right)\\
&=\sqrt5\sum_{k=0}^\infty\frac{(-1)^k}{\phi^{2k+1}-(-1)^k}
\end{align}
$$
Since $\frac{\sqrt5}{\phi^{19}+1}=0.0002392$,
$$
\sqrt5\sum_{k=0}^8\frac{(-1)^k}{\phi^{2k+1}-(-1)^k}=3.3600587
$$
is less than $0.0002392$ too high.