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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-Level Student
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Franklin Pezzuti Dyer
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  • program – Kenny Lau Dec 25 '17 at 23:31
  • Have you tried using Binet’s formula for $F_{n}$? – Nah Dec 25 '17 at 23:51
  • Inverse symbolic calculator finds nothing. – Carl Schildkraut Dec 25 '17 at 23:58
  • @MatíasBruna Yes, but it doesn't help at all. The series isn't geometric due to addition in the denominator. – Franklin Pezzuti Dyer Dec 26 '17 at 00:05
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    Series of the $\sum_{n\geq 1}\frac{(-1)^n}{F_n F_{n+k}}$ kind are simple to compute since they are related to the convergents of the golden ratio, but we cannot state the same if we remove the alternating sign. In any case, it is pretty simple to evaluate such sums numerically by ad-hoc acceleration techniques, since $F_n\approx \frac{\varphi^n}{\sqrt{5}}$, so they all are approximately geometric (the sums in your previous question, too). – Jack D'Aurizio Dec 26 '17 at 00:15
  • @JackD'Aurizio Well, I have computed a closed form for the non-alternating variation with even $k$, which is why I am curious about the case for odd $k$. I will edit the closed-form into my question. – Franklin Pezzuti Dyer Dec 26 '17 at 14:04
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    You might look at the Fibonacci Quarterly – marty cohen Dec 26 '17 at 14:48
  • Your problem reminds me of prof. Vernon Hoggatt. It was said that he suicide on a clothes line in a peculiar mathematical manner. However, every time I encountered him, he was the most cheerful guy imaginable. I am sorry that I never took a class from him. Fibonacci theory would be a lot farther along if he had been with us a few more years. – richard1941 Dec 27 '17 at 03:07
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    Not sure if this helps at all, but from A290565 $\sum_{k=1}^\infty \frac{1}{F_kF_{k+1}}=2\left(\sum_{k=1}^\infty \frac1{\phi^k F_k}\right) - \tfrac1\phi$. – g.kov Apr 05 '21 at 14:54
  • Looks like: $$\Phi_n(x)=\sum_{k=1}^\infty \frac{x^{k+1}}{F_kF_{k+n}}=\sum _{k=1}^{\infty } \frac{5 \left(-\frac{1}{\phi }\right)^{-2 k-n} x^{1+k}}{\left(-1+(-1)^k \phi ^{2 k}\right) \left(-1+(-1)^{k+n} \phi ^{2 (k+n)}\right)}$$ – Mariusz Iwaniuk Apr 05 '21 at 16:45
  • How did you find $\Phi_1(-1)$ please? @g.kov do you know how he found it? – A-Level Student Apr 10 '21 at 21:49
  • @A-Level Student: Check out tis one: A265290, hope this could help. – g.kov Apr 11 '21 at 02:49

1 Answers1

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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.)

user90369
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