Summing up the reciprocals of Fibonacci numbers

This is the Reciprocal Fibonacci Constant, and the closed form of this constant is

$\dfrac { \sqrt { 5 } }{ 4 } \left( \dfrac { 1 }{ 2Log\left( \phi \right) } \left( Log\left( 5 \right) +2{ \psi }_{ { \phi }^{ -4 } }\left( 1 \right) -4{ \psi }_{ { \phi }^{ -2 } }\left( 1 \right) \right) \quad +{ \quad \left( \vartheta _{ 2 }\left( 0,{ \phi }^{ -2 } \right) \right) }^{ 2 } \right) =3.3598856662431775532...$

where $\phi$ is the golden ratio , $\psi$ is the q-Polygamma function , and $\vartheta$ is the elliptic theta function

It's much easier to just add up the first 30 terms or so to get 5 decimal point accuracy.

With $\phi= \tfrac12(1+\sqrt{5})$ , we have $F_n = \tfrac{1}{\sqrt{5}}\big(\phi^n - (-\phi^{-1})^n\big)$ , and hence $\frac{1}{F_n} \; = \; \frac{\sqrt{5}}{\phi^n -(-\phi^{-1})^n} \; = \; \frac{\sqrt{5}}{\phi^n}\big(1 - (-\phi^{-2})^n\big)^{-1} \; = \; \sqrt{5}\sum_{k \ge 0}(-1)^{kn} \phi^{-(2k+1)n}$ so that $\sum_{n=1}^\infty \frac{1}{F_n} \; = \;\sqrt{5}\sum_{k = 0}^\infty \sum_{n=1}^\infty (-1)^{kn}\phi^{-(2k+1)n} \; = \; \sqrt{5}\sum_{k=0}^\infty \frac{(-1)^k}{\phi^{2k+1} - (-1)^k}$ This series is alternating, and its summands converge to $0$ exponentially, so we have rapid convergence. The approximation $\sqrt{5}\sum_{k=0}^{12} \frac{(-1)^k}{\phi^{2k+1} - (-1)^k} \approx \boxed{3.35989}$ gives the sum to $5$ DP accuracy.