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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? $$\Phi_1(1)=\sum_{k=1}^\infty \frac{1}{F_kF_{k+1}}$$

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  • $\begingroup$ program $\endgroup$ – Kenny Lau Dec 25 '17 at 23:31
  • $\begingroup$ Have you tried using Binet’s formula for $F_{n}$? $\endgroup$ – Matías Bruna Dec 25 '17 at 23:51
  • $\begingroup$ Inverse symbolic calculator finds nothing. $\endgroup$ – Carl Schildkraut Dec 25 '17 at 23:58
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    $\begingroup$ 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). $\endgroup$ – Jack D'Aurizio Dec 26 '17 at 0:15
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    $\begingroup$ You might look at the Fibonacci Quarterly $\endgroup$ – marty cohen Dec 26 '17 at 14:48
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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.)

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