# Is there a closed form of $\displaystyle \sum_{k=0}^{\infty}{\frac{\phi^{xk}}{k!_F}}$

where $$\phi = \frac{1+\sqrt{5}}{2}$$ and $$k!_F$$ is the fibonorial of $$k$$, or the product of the first $$k$$ Fibonacci numbers? My hunch is that, this can be represented as a function in terms of the constant in the case where x=1, so I tried the following: $$\begin{split} S = \sum_{k=0}^{\infty}{\frac{\phi^k}{k!_F}} & =\sum_{k=0}^{\infty}{\frac{F_k\phi+F_{k-1}}{k!_F}} \\ &=\sum_{k=0}^{\infty}\Bigg (\phi\frac{F_{k}}{k!_F}+\frac{F_{k-1}}{k!_F}\Bigg)\\ & =\phi \displaystyle \sum_{k=0}^{\infty}\frac{F_{k}}{k!_F}+\sum_{k=0}^{\infty}\frac{F_{k-1}}{k!_F} \end{split}$$ How should I continue from here?. I'm asking for a closed form expression but any alternate representations as an integral or another series are greatly appreciated as well.

(Fibonacci numbers as defined by the recurence $$F_k = F_{k-1}+ F_{k-2}$$ and $$F_0=F_1=1$$)

(Edit: The first couple of terms of $$\displaystyle \sum_{k=0}^{\infty}\frac{1}{(k-2)!_F}\frac{1}{F_k}$$ and the first term of $$\displaystyle \sum_{k=0}^{\infty}\frac{1}{(k-1)!_F}$$ aren't defined so I changed it back to $$\displaystyle \sum_{k=0}^{\infty}\frac{F_{k-1}}{k!_F}$$ and $$\displaystyle \sum_{k=0}^{\infty}\frac{F_{k}}{k!_F}$$ respectively )

(Edit: It seems that $$\displaystyle \sum_{k=0}^{\infty} \frac{F_k}{k!_F} = \sum_{k=0}^{\infty} \frac{1}{k!_F}$$)

• I think it is always a good idea to give your definition of Fibonacci numbers: There are two natural indexations: $F_0=0,F_1=1$ or $F_0=F_1=1$. I guess you want the second one. Aug 21, 2021 at 16:11
• The first term in the last summation is ${1\over(-2)!_F}{1\over F_0}$, and I'm having trouble seeing what $(-2)!_F$ could mean. If you start at $k=2$, you get a rapidly converging series, so you could compute a few decimals and then see whether you can conjecture a closed form for the result. Aug 21, 2021 at 23:12
• It's not equivalent, because it has terms that don't make sense. I think you have to split of the terms with $k=0$ and $k=1$. Aug 22, 2021 at 8:53
• Right, but me simplifying $\displaystyle \sum_{k=0}^{\infty} \frac{F_{k-1}}{k!_F}$ to $\displaystyle \sum_{k=0}^{\infty} \frac{1}{(k-2)!_F}\frac{1}{F_k}$ was probably a mistake. @GerryMyerson Aug 22, 2021 at 9:00
• $\sum_0{F_k\over k!_F}=1+\sum_1{F_k\over k!_F}=1+\sum_1{1\over(k-1)!_F}=1+\sum_0{1\over k!_F}$. Aug 22, 2021 at 12:51