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?
On
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 $$S=\sqrt5\sum_{n=1}^\infty\left(\frac1{\beta^{-n}-1}-\frac1{(\beta^2)^{-n}-1}\right)$$ $$=\sqrt5\left(\frac{\psi_\beta(1)+\ln(1-\beta)}{\ln\beta}-\frac{\psi_{\beta^2}(1)+\ln(1-\beta^2)}{2\ln\beta}\right)$$ $$=\frac{\sqrt5}{2\ln\beta}\left(2\psi_\beta(1)-\psi_{\beta^2}(1)+\ln\frac{1-\beta}{1+\beta}\right)$$
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 $$\left( \begin{array}{cc} 1 & 0.389061423334175 \\ 2 & 0.389082999708164 \\ 3 & 0.389083066686468 \\ 4 & 0.389083066894475 \\ 5 & 0.389083066895121 \\ 6 & 0.389083066895123 \end{array} \right)$$