<s>[INST] Finding relatives of the series $\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}$. 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.
 [/INST] Yes, you are right. There is a family. A little experimentation shows that,
$$\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}$$
$$\varphi =\frac{13}{2^3}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}$$
$$\varphi =\frac{207}{2^7}+\sum_{k=0}^{\infty}(-1)^{k+2}\frac{C_{k+2}}{2^{4k+11}}$$
$$\varphi =\frac{1657}{2^{10}}+\sum_{k=0}^{\infty}(-1)^{k+3}\frac{C_{k+3}}{2^{4k+15}}$$
$$\varphi =\frac{53019}{2^{15}}+\sum_{k=0}^{\infty}(-1)^{k+4}\frac{C_{k+4}}{2^{4k+19}}$$
and so on. Sorry, but the numerators of the fractions are not Fibonacci numbers. It seems hard to predict what they will be.
</s>