Fibonacci sum: $\sum\limits_{k\ge0}\frac{F_{2k+1}}{2k+1}\left(\frac{2+2\sqrt{2}}{1+\sqrt{\frac{17+8\sqrt{2}}{5}}}\right)^{2k+1}(-\frac{1}{5})^k$ 
Prove $$\frac{3\pi}{8}=\sum\limits_{k\ge0}\left(-\frac{1}{5}\right)^k\frac{F_{2k+1}}{2k+1}q^{2k+1}$$
  where $q=\frac{2+2\sqrt{2}}{1+\sqrt{\frac{17+8\sqrt{2}}{5}}}$, and $F_n$ are the Fibonacci numbers. This result is from Wolfram.

My ideas as to a proof are limited. I know that this is some sort of series for the inverse tangent because $3\pi/8=\arctan(1+\sqrt{2})$, but I have never seen any series for $\arctan$ involving the Fibonacci numbers. Essentially, it comes to the explicit evaluation of the function $$f(x)=\sum_{k\ge0}\frac{F_{2k+1}}{2k+1}x^{k}.$$
Indeed, the sum in question is the value $qf(-q^2/5)$.
Potentially connected is the closed form
$$\sum_{k\ge0}F_kx^k=\frac{x}{1-x-x^2},$$
perhaps we could get some sort of $\arctan$-related integral out of this.
Could I have some help evaluating $f$? Thanks.
 A: Let $$f(x) = \sum_{k\geq 0}F_{k+1}x^k$$ so that $$(1-x-x^2)f(x)=F_1+(-F_1+F_2)x$$ or $$f(x) =\frac{1}{1-x-x^2}$$ Integrating the above we get $$F(x) =\int_{0}^{x}\frac{dt}{1-t-t^2}=\sum_{k\geq 0}\frac{F_{k+1}}{k+1}x^{k+1}$$ The integrand on left side can be written $$\frac{1}{(1-at)(1-bt)}=\frac{A}{1-at}+\frac{B}{1-bt}$$ where $a, b$ are roots of $z^2-z-1=0$ and $$A=\frac{a} {a-b}, B=-\frac{b} {a-b} $$ Letting $$a=\frac{1+\sqrt{5}}{2},b=\frac{1-\sqrt {5}}{2}$$ we get $a-b=\sqrt{5}$ and hence we have $$F(x) =\frac{1}{\sqrt {5}}\log\frac{1-bx}{1-ax}$$ and $$G(x) =F(x) - F(-x) =2\sum_{k\geq 0}\frac{F_{2k+1}}{2k+1}x^{2k+1}$$ Replacing $x$ by $ix$ we get $$\sum_{k\geq 0}(-1)^{k}\frac{F_{2k+1}}{2k+1}x^{2k+1}=-i\frac{G(ix)} {2}$$ The sum in question is $$-i\sqrt {5}\frac{G(iq/\sqrt{5})}{2}$$ Now $$G(x) =\frac{1}{\sqrt{5}}\log\frac{1+\sqrt {5}x+x^2}{1-\sqrt{5}x+x^2}$$ and hence $$G(iq/\sqrt{5})=\frac{1}{\sqrt{5}}\log\frac{5+5iq-q^2}{5-5iq-q^2}=\frac{2i}{\sqrt {5}}\arctan\frac{5q}{5-q^2}$$ The desired sum is thus equal to $$\arctan\frac{5q}{5-q^2}$$ The fact that $5q/(5-q^2)=1+\sqrt{2}$ needs some effort and I have not been able to do so efficiently.

As pointed out in comments the series for $G$ diverges if $|x|>1/\phi$ or equivalently $|q| >\sqrt{5}/\phi$. The value of $q$ used in question is greater than $\sqrt{5}/\phi$ and hence the sum in question diverges.
