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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?

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  • $\begingroup$ Mathematica says $0.389083$. $\endgroup$ – Kenta S Jul 3 at 4:38
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    $\begingroup$ Is there a reason to believe it has a closed form? $\endgroup$ – runway44 Jul 3 at 4:39
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    $\begingroup$ It is hard to explain in comment, but I think this paper of Melham and Shannon may be helpful for you. See the formula 2.9 $\endgroup$ – GAVD Jul 3 at 4:46
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    $\begingroup$ Tough question could we know the source, or inspiration? $\endgroup$ – Anirudh Jul 3 at 4:47
  • $\begingroup$ @runway44 Yes, there is a closed form, if you allow some special functions. See my answer. $\endgroup$ – Parcly Taxel Jul 3 at 5:36
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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 $$S=\sqrt5\sum_{n=1}^\infty\left(\frac1{\beta^{-n}-1}-\frac1{(\beta^2)^{-n}-1}\right)$$ $$=\sqrt5\left(\frac{\psi_\beta(1)+\ln(1-\beta)}{\ln\beta}-\frac{\psi_{\beta^2}(1)+\ln(1-\beta^2)}{2\ln\beta}\right)$$ $$=\frac{\sqrt5}{2\ln\beta}\left(2\psi_\beta(1)-\psi_{\beta^2}(1)+\ln\frac{1-\beta}{1+\beta}\right)$$

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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 $$\left( \begin{array}{cc} 1 & 0.389061423334175 \\ 2 & 0.389082999708164 \\ 3 & 0.389083066686468 \\ 4 & 0.389083066894475 \\ 5 & 0.389083066895121 \\ 6 & 0.389083066895123 \end{array} \right)$$

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  • $\begingroup$ What does $S_p$ mean? $\endgroup$ – Marian G. Jul 3 at 5:14
  • $\begingroup$ @MarianG.. Compute exactly the sum for the first $p$ terms and, for $n>p$, use Binet formula with the approximation. $\endgroup$ – Claude Leibovici Jul 3 at 5:22
  • $\begingroup$ I was confused since $S_p$ could also denote the $p$-th partial sum. Thanks. $\endgroup$ – Marian G. Jul 3 at 5:37

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