What is the exact value of $\sum\limits_{n=1}^\infty \frac{1}{F_n}$? $F_n$ denotes the $n^{th}$ Fibonacci number.
Wolframalpha gave me this answer: $$\sum_{n=1}^{\infty}\frac{1}{F_n}\ =\frac{1}{4}\sqrt{5}\left(\frac{2\psi_{\frac{1}{\phi^{4}}}^{0}\left(1\right)-4\psi_{\frac{1}{\phi^{2}}}^{0}\left(1\right)+\ln\left(5\right)}{2\ln\left(\frac{1}{2}\left(1+\sqrt{5}\right)\right)}+ϑ_{2}\left(0,\frac{1}{\phi^{2}}\right)^{2}\right)\approx3.35989$$ $ϑ_\alpha(x,q)$ is the theta function
$\psi_q(z)$ gives the q-digamma function
$\phi$ is the golden ratio
I'm not really sure how it got this solution and I don't really know where to begin solving this one. Also, is there a more concise way to write the solution?
Edit: After simplifying, I got: $$\sum_{n=1}^{\infty}\frac{1}{F_n}=\sum_{n=1}^{\infty}\frac{2\sqrt{5}}{\left(1+\sqrt{5}\right)^{n}-\left(1-\sqrt{5}\right)^{n}}$$