Series involving Fibonacci Numbers: $\sum_{k=1}^\infty \frac{1}{F_kF_{k+1}}$ I will start my question with a bit of information that I think may be helpful to potential answerers. If you don't want to read it, skip down to the question.
BACKGROUND:
I'm investigating series in the form
$$\Phi_n(x):=\sum_{k=1}^\infty \frac{x^{k+1}}{F_kF_{k+n}}$$
for $x=\pm 1$ ($F_k$ is the Fibonacci sequence with $F_1=F_2=1$). I have easily calculated
$$\Phi_2 (1)=1$$
and somewhat less-easily calculated
$$\Phi_4(1)=\frac{7}{18}$$
using telescoping, and I can calculate $\Phi_n(-1)$ for any positive integer $n$ because I have found
$$\Phi_1(-1)=\frac{1}{\phi}$$
and I have discovered the recurrence
$$\Phi_n(-1)=\frac{1}{F_n\phi}-\frac{F_{n-1}}{F_n^2}+\frac{F_{n-1}}{F_n}\Phi_{n-1}(-1)$$
for $n\ge 2$. I can also calculate $\Phi_n(1)$ for even $n$ because of the recurrences
$$\Phi_n(1)=\frac{\Phi_1(1)-F_{n-1}\Phi_{n-1}(1)}{F_n}+\frac{F_{n-1}}{F_n^2}$$
and
$$\Phi_n(1)=\frac{F_{n-2}}{F_n}\Phi_{n-2}(1)-\frac{F_{n-2}}{F_n F_{n-1}}+\frac{F_{n-1}}{F_n^2}$$
However, I cannot figure out how to calculate
$$\Phi_1(1)=\sum_{k=1}^\infty \frac{1}{F_kF_{k+1}}$$
NEW INFORMATION:
I have computed closed-form expressions for $\Phi_n(-1)$ and $\Phi_{2n}(1)$ for positive integer $n$:
$$\Phi_n (-1)=\frac{n}{F_n\phi}-\frac{1}{F_n}\sum_{k=1}^{n-1}\frac{F_k}{F_{k+1}}$$
$$\Phi_{2n} (1)=\frac{1}{F_{2n}}+\frac{1}{F_{2n}}\sum_{k=1}^{n-1}\frac{1}{F_{2k+1}F_{2k+2}}$$
I have also discovered a more general recurrence relating $\Phi_n(x)$ to $\Phi_{n-1}(x)$:
$$\Phi_n(x)=\frac{\Phi_1(x)}{F_n}-\frac{F_{n-1}}{F_n}\frac{\Phi_{n-1}(x)}{x}+\frac{F_{n-1}x}{F_n^2}$$
QUESTION:
How can I calculate the value of this series?
$$\Phi_1(1)=\sum_{k=1}^\infty \frac{1}{F_kF_{k+1}}$$
 A: A note.
The recurrence $\enspace\displaystyle \Phi_n(x)=\frac{\Phi_1(x)}{F_n}-\frac{F_{n-1}}{F_n}\frac{\Phi_{n-1}(x)}{x}+\frac{F_{n-1}x}{F_n^2}\,$ multiplicated with $\,F_n\enspace$ and 
$\,n\to\infty\,$ leads to $\,\displaystyle \Phi_1(x)=-\frac{2x}{1+\sqrt{5}}+\sqrt{5}(1+\frac{1}{x})\sum\limits_{k=1}^\infty \frac{x^{k+1}}{(\frac{3+\sqrt{5}}{2})^k-(-1)^k}\enspace$ and therefore  
to $\enspace\displaystyle \Phi_1(1)=-\frac{2}{1+\sqrt{5}}+2\sqrt{5}\, f(\frac{3+\sqrt{5}}{2})\enspace$ with $\enspace\displaystyle f(x):=\sum\limits_{k=1}^\infty \frac{1}{x^k-(-1)^k} \,$ , $\enspace |x|>1\,$. 
It remains the problem to simplify $f(x)$ which is independend of the Fibonacci numbers.
$\displaystyle f(-\frac{1}{x}) = x\frac{d}{dx}\ln g(x) \enspace$ for $\enspace g(x):=\prod\limits_{k=1}^\infty (1-x^k)^{\frac{(-1)^{k-1}}{k}}\,$ with $\,|x|<1\,$ . 
I don't know if $\,g(x)\,$ is easier or worse to discuss than Euler's pentagonal number theorem, based on the Jacobi triple product (e.g. https://en.wikipedia.org/wiki/Jacobi_triple_product).
Another possibility is to discuss $\enspace\displaystyle h(x):=\sum\limits_{k=1}^\infty\frac{x^k}{F_k}\enspace$ for $\,\displaystyle |x|<\frac{1+\sqrt{5}}{2}\enspace$ because of  
$\,\displaystyle \Phi_1(x)=-\frac{2x}{1+\sqrt{5}}+(1+x)\,h(\frac{2x}{1+\sqrt{5}})\,$ . $\,$ Sorry, I haven't seen any literature about $\,h(x)\,$ .
About the special case, the irrational reciprocal Fibonacci constant $\,\psi=h(1)\,$ (https://en.wikipedia.org/wiki/Fibonacci_number), is said that there is no closed formula known.
I think we can assume, that there is no closed formula for $\,\Phi_1(x)\,$ . (But maybe it's possible to find an integral for that function which would be interesting.)
