$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$? I stumble upon the expression
$$
\sum_{k=0}^\infty 
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}
$$
and it seems to converge to
$$
\frac{3-\sqrt{5}}{2}
$$
Do they equate ? How to prove that ?
Not sure if it's helpful :
$\frac{1}{k+1}\binom{3k+1}{k} , \; k\ge0$ is the OEIS sequence $A006013$.
I only have fundamental knowledge in combinatorics so I could only check it numerically that the convergence seems to be true . However I wouldn't mind learning new theories   .

Let
$$S(z)=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}(1-z)^{2k+1}.$$
One has
\begin{align}S(z)&=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}\sum_{l=0}^{2k+1}(-1)^l{{2k+1}\choose{l}}z^l \\
&=\sum_{n=0}^\infty\sum_{k+l=n}(-1)^l{{3k+1}\choose{k}}{{2k+1}\choose{l}}z^n\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}{{3k+1}\choose{k}}{{2k+1}\choose{n-k}}z^n
\end{align}
On the other hand,
$${{3k+1}\choose{k}}{{2k+1}\choose{n-k}}=\frac{(3k+1)!}{k!(2k+1)!} \frac{(2k+1)!}{(n-k)!(3k+1-n)!}={n\choose k}{3k+1\choose n}.$$
Moreover, we have the following combinatorial identity:
$$\sum_{k=0}^n(-1)^{n-k}{n\choose k}{3k+1\choose n}=3^n.$$
One way to verify this identity is to write
$$f(x)=x(1+x^3)^n=\sum_{k=0}^n {n\choose k}x^{3k+1}.$$
and notice that the $n$th derivative of $x(1+x^3)^n$ at $x=-1$ is $-3^nn!$, while the $n$th derivative of $\sum_{k=0}^n {n\choose k}x^{3k+1}$ is $\sum_{k=0}^n n!{3k+1\choose n}{n\choose k}x^{3k+1-n}$.
Combining these, we deduce that
$$S(z)=\sum_{n=0}^\infty 3^nz^n=\frac{1}{1-3z}.$$
Next, let
$$G(u)=\sum_{k=0}^\infty \frac{1}{k+1}{3k+1\choose k}u^{k+1},$$
and $u=z(1-z)^2$. Then
$$\frac{dG}{dz}=\frac{dG}{du}\frac{du}{dz}.$$
Since $dG/du=\sum_{k=0}^\infty {3k+1 \choose k}(z(1-z)^2)^k=\frac{S(z)}{1-z}=\frac{1}{(1-z)(1-3z)}$ and $du/dz=1-4z+3z^2$, we have $\frac{dG}{dz}=1$, and so $G(z)=z$ (note that $G(0)=0$).
We have proved that
$$\sum_{k=0}^\infty \frac{1}{k+1}{3k+1 \choose k}(z(1-z)^2)^{k+1}=z.$$
Finally, let $z=(3-\sqrt{5})/2$, for which $z(1-z)^2=1/8$, to conclude that
$$\sum_{k=0}^\infty \frac{1}{k+1}{3k+1 \choose k}\frac{1}{2^{3k+3}}=\frac{3-\sqrt{5}}{4}.$$
