Sum of reciprocals of Fibonacci numbers convergence

You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:

$$ S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$ such that: $$ \frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}. $$

For $n \ge 3$, we have

$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}} = \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n} = \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\ = \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n} $$

The partial sums is monotonic increasing and bounded from above. $$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n} \le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$ As a result, the series converges.

Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$