Let $F_n$ be the $n^\text{th}$ Fibonacci number. I wanted to calculate $$\sum_{n\ge1}\frac{1}{F_n}$$ I simplified it to $$\sqrt{5}\sum_{n\ge1}\frac{1}{\varphi^n-\phi^n}$$ But this didn't seem to help. The value is approximately $3.35988566624317755317201130291892...$. The OEIS entry of this constant is this. Mathematica gives the answer $$\frac{\sqrt{5}}{4} \left(\frac{\log 5+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)}{2\log\phi}+\vartheta_{2}^{2}\left(\phi^{-2}\right)\right)$$ Where $\vartheta$ is the Elliptic Theta function and $\psi$ is the q-polygamma function. This looks very hard to derive. But, this can be calculated, which means that there is surely a method to do this. So, my question is,
How can the closed form be derived?
If the derivation is very long, an article containing the proof can be mentioned, but a derivation in the answers would be more helpful.
$$F_n=\frac{1}{\sqrt{5} }\left({\varphi^n-\phi^n}\right)=\frac{\varphi^n}{\sqrt{5} }\left(1-\left(\frac{\phi }{\varphi }\right)^n\right)$$ $$\frac 1{F_n}=\frac{\sqrt{5} }{\varphi^n} \frac 1{1-\left(\frac{\phi }{\varphi }\right)^n }$$
$$\sum_{n=1}^{\infty}\frac{1}{F_n}=\sqrt{5}\sum_{n=1}^{\infty}\frac{1 }{\varphi^n} \frac 1{1-\left(\frac{\phi }{\varphi }\right)^n }=\sqrt{5}\sum_{n=1}^{\infty}\frac{1 }{\varphi^n}\sum_{k=0}^{\infty}\left(\frac{\phi }{\varphi }\right)^{k n}$$
It is now for the second portion of the summand where the special functions are comming from.
You can have very good approximations writing $$\sum_{n=1}^{\infty}\frac{1}{F_n}\sim \sum_{n=1}^{p}\frac{1}{F_n}+\sqrt{5}\sum_{n=p+1}^{\infty}\frac{1 }{\varphi^n}=\sum_{n=1}^{p}\frac{1}{F_n}+\frac{5+3 \sqrt{5}}{2} \varphi ^{-(p+1)} $$ $$\left( \begin{array}{cc} p & \text{approximation} \\ 5 & 3.35957125458259727076544117389 \\ 10 & 3.35988589577114994666863433486 \\ 15 & 3.35988566607489618438018003508 \\ 20 & 3.35988566624330092650853900324 \\ 25 & 3.35988566624317746272239950104 \\ 30 & 3.35988566624317755323832329939 \\ 35 & 3.35988566624317755317196268711 \\ 40 & 3.35988566624317755317201133856 \\ 45 & 3.35988566624317755317201130289 \\ 50 & 3.35988566624317755317201130292 \end{array} \right)$$