How to evaluate the series $\sum_{n=1}^\infty \frac{1}{F_{4n}}$? Where $F_1=1$, $F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ How would I go about evaluating the following series: $$\sum_{n=1}^\infty \frac{1}{F_{4n}}$$ 
Where $F_1=1$, $F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n \ge 3$
Not sure how to go about this except maybe by using the close form for the nth fibonacci number, but that seems like way too much arithmetic. Any ideas?
 A: If you allow what Mathworld calls $q$-polygamma functions, there is a closed form for the series. We start by writing the closed form of $F_{4n}$:
$$F_{4n}=\frac{(7+3\sqrt5)^n-(7-3\sqrt5)^n}{2^n\sqrt5}$$
Thus the series becomes
$$S=\sum_{n=1}^\infty\frac{2^n\sqrt5}{(7+3\sqrt5)^n-(7-3\sqrt5)^n}$$
$$=\sqrt5\sum_{n=1}^\infty\frac{\left(\frac2{7-3\sqrt5}\right)^n}{\left(\frac{7+3\sqrt5}{7-3\sqrt5}\right)^n-1}$$
Define $\alpha=\frac2{7-3\sqrt5}$, then notice that
$$\alpha^2=\frac{7+3\sqrt5}{7-3\sqrt5}$$
Thus we can write
$$S=\sqrt5\sum_{n=1}^\infty\frac{\alpha^n}{\alpha^{2n}-1}$$
$$=\sqrt5\sum_{n=1}^\infty\left(\frac1{\alpha^n-1}-\frac1{\alpha^{2n}-1}\right)$$
Now define $\beta=\frac1\alpha=\frac{7-3\sqrt5}2$, and use the $q$-polygamma relation
$$\sum_{n=1}^\infty\frac1{x^{-n}-1}=\frac{\psi_x(1)+\ln(1-x)}{\ln x}$$
to derive
$$S=\sqrt5\sum_{n=1}^\infty\left(\frac1{\beta^{-n}-1}-\frac1{(\beta^2)^{-n}-1}\right)$$
$$=\sqrt5\left(\frac{\psi_\beta(1)+\ln(1-\beta)}{\ln\beta}-\frac{\psi_{\beta^2}(1)+\ln(1-\beta^2)}{2\ln\beta}\right)$$
$$=\frac{\sqrt5}{2\ln\beta}\left(2\psi_\beta(1)-\psi_{\beta^2}(1)+\ln\frac{1-\beta}{1+\beta}\right)$$
A: I do not know if there is a closed form formula but you can have a good approximation of it writing
$$S=\sum_{n=1}^\infty \frac{1}{F_{4n}}=\sum_{n=1}^p \frac{1}{F_{4n}}+\sum_{n=p+1}^\infty \frac{1}{F_{4n}}$$ and use, after some $p$, $F_k\sim \frac{\phi^k}{\sqrt 5}$ (Binet formula). This means
$$S_p=\sum_{n=1}^p \frac{1}{F_{4n}}+\sqrt{5}\frac{ \phi ^{-4 p}}{\phi ^4-1}$$ which converges very fast as shown below
$$\left(
\begin{array}{cc}
 1 & 0.389061423334175 \\
 2 & 0.389082999708164 \\
 3 & 0.389083066686468 \\
 4 & 0.389083066894475 \\
 5 & 0.389083066895121 \\
 6 & 0.389083066895123 
\end{array}
\right)$$
