TITLE: Finding relatives of the series $\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}$.
QUESTION [7 upvotes]: Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$
$$
\begin{align*}
\varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\
\varphi &=\frac{2}{1}+\sum_{k=0}^{\infty}\cdots&(2)\\
\varphi &=\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}&(3)\\
\varphi &=\frac{5}{3}+\sum_{k=0}^{\infty}\cdots&(4)\\
\varphi &=\frac{8}{5}+\sum_{k=0}^{\infty}\cdots&(5)\\
\varphi &=\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2(k+1))!}{((k+1)+1)!(k+1)!2^{4(k+1)+3}}&(6)\\
\vdots&\\
\end{align*}
$$
When looking at the leading terms of $(3)$ and $(6)$  $\;\frac{3}{2}$ and $\frac{13}{8}$ respectively, one is tempted to conjecture that there are similar formulas to fill the holes in the above table.
I'd like to know if such family of formulas exist.
Thanks.

EDIT:
Note that both formulas connect the Golden Ratio $\varphi$ to Catalan Numbers
$$
C_{k}=\frac{(2k)!}{(k+1)!k!}
$$
so for $(3)$ we have
$$
\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}
$$
and for $(6)$ we have
$$
\varphi =\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4(k+1)+3}}
$$
So, maybe this could be used, somehow, to find the other formulas.

REPLY [5 votes]: Let us take the first term out of the summation in equation $(3)$.
$$\begin{align}
\varphi&=\frac{3}{2}+\sum_{k=0}^\infty (-1)^k\frac{C_k}{2^{4k+3}}\\
&=\frac{3}{2}+\frac{1}{8}+\sum_{k=1}^\infty (-1)^k\frac{C_k}{2^{4k+3}}\\
&=\frac{13}{8}+\sum_{k=0}^\infty (-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}\\
\end{align}$$
The result is equation $(6)$.
Iterating the same procedure gives all formulas listed in Tito Piezas' answer, so this family is obtained directly from equation $(3)$.
The general formula for the fractions is therefore
$$\frac{3}{2}+\sum_{k=0}^N (-1)^{k}\frac{C_k}{2^{4k+3}}$$
Egyptian fraction versions of $(2)$, $(4)$ and $(6)$ are given by
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$
$$\varphi=\frac{5}{3}-\sum_{k=0}^\infty \frac{1}{F(2^{k+3})}$$
$$\varphi=\frac{13}{8}-\sum_{k=0}^\infty \frac{1}{F(3·2^{k+2})}$$
and additional series involving Fibonacci numbers are given by
$$\varphi=\frac{34}{21}-\sum_{k=0}^\infty \frac{1}{F(2^{k+4})}$$
$$\varphi=\frac{89}{55}-\sum_{k=0}^\infty \frac{1}{F(5·2^{k+2})}$$
$$\varphi=\frac{233}{144}-\sum_{k=0}^\infty \frac{1}{F(3·2^{k+3})}$$
$$\varphi=\frac{610}{377}-\sum_{k=0}^\infty \frac{1}{F(7·2^{k+2})}$$
The general pattern is

$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$

Alternatively, we can write formulas (1), (2), (4) and (5) as a linear combination of formulas (3) and (6), from relationships 
$$1=5\left(\frac{3}{2}\right)-4\left(\frac{13}{8}\right)$$
$$1=-3\left(\frac{3}{2}\right)+4\left(\frac{13}{8}\right)$$
$$\frac{5}{3}=-\frac{1}{3}\left(\frac{3}{2}\right)+\frac{4}{3}\left(\frac{13}{8}\right)$$
$$\frac{8}{5}=\frac{1}{5}\left(\frac{3}{2}\right)+\frac{4}{5}\left(\frac{13}{8}\right)$$
After some algebra, the results are

$$\begin{align}\varphi &= 1 + 3\sum_{k=0}^\infty \frac{(-1)^k(2k)!(4k+7)}{k!(k+2)!16^{k+1}}\tag{1}\\
\\
\varphi &= 2 - \sum_{k=0}^\infty \frac{(-1)^k(2k)!(8k+13)}{k!(k+2)!16^{k+1}}\tag{2}\\
\\
\varphi &= \frac{5}{3}-\frac{1}{3}\sum_{k=0}^\infty \frac{(-1)^k(2k)!(4k+5)}{k!(k+2)!16^{k+1}}\tag{4}\\
\\
\varphi &= \frac{8}{5}+\frac{3}{5}\sum_{k=0}^\infty \frac{(-1)^k(2k)!}{k!(k+2)!16^{k+1}}\tag{5}
\end{align}$$

Two observations: the procedure used has little to do with these particular fractions, similar results could be expected for any fraction; the simplest formula, (5), is obtained from a true interpolation, with both coefficients positive.
Since all these fractions lie between $1$ and $2$, we may take equations (1) and (2) as a basis and parameterize the family of approximations $\varphi\approx \frac{p}{q}$ according to
$$\frac{p}{q}=2\alpha+1(1-\alpha)=\alpha + 1$$
so the general equation may be written
$$\varphi=\frac{p}{q}+\sum_{k=0}^\infty \frac{(-1)^k(2k)!((3-5\alpha)4k+21-34\alpha)}{k!(k+2)!16^{k+1}}$$
where $\alpha=\dfrac{p}{q}-1$.
When $\dfrac{p}{q}=\dfrac{3}{2}$ then $\alpha=\dfrac{1}{2}$ and the multiplying polynomial $(12-20\alpha)k+21-34\alpha$ reduces to $(2k+4)=2(k+2)$. This cancels the largest factor of $(k+2)$! in the denominator and equation (3) is obtained.