$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ 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   .

Some thoughts:
We use the integral representation
$$\binom{3k + 1}{k} = \frac{1}{2\pi}\int_{-\pi}^\pi (1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^{3k + 1}(2^{-1}\mathrm{e}^{\mathrm{i}t})^{-k}\mathrm{d} t. \tag{1}$$
(Note: Similar to https://functions.wolfram.com/GammaBetaErf/Binomial/07/02/)
We have
\begin{align*}
 &\sum_{k=0}^\infty 
 \frac{1}{k+1}
 \binom{3k+1}{k}
 \left( \frac{1}{2} \right)^{3k+2}\\[6pt]
 =\,& \frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=0}^\infty \frac{1}{k + 1}2^{-(3k + 2)}(1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^{3k + 1}(2^{-1}\mathrm{e}^{\mathrm{i}t})^{-k}\, \mathrm{d} t\\[6pt]
 =\,& \frac{1}{2\pi}\int_{-\pi}^\pi
 \frac{1 + z}{4}\sum_{k=0}^\infty \frac{1}{k + 1} \left(\frac{(1 + z)^3}{8z}\right)^k\Big\vert_{z = 2^{-1}\mathrm{e}^{\mathrm{i}t}}\, \mathrm{d} t \\[6pt]
 =\,& \frac{1}{2\pi}\int_{-\pi}^\pi
 \frac{-2z}{(1 + z)^2}
 \ln \left(1 - \frac{(1 + z)^3}{8z} \right)\Big\vert_{z = 2^{-1}\mathrm{e}^{\mathrm{i}t}}\,\mathrm{d} t \tag{2}\\[6pt]
 =\,& \frac{3 - \sqrt 5}{2} \tag{3}
\end{align*}
where we have used
$$\sum_{k=0}^\infty \frac{1}{k + 1} a^k = - \frac{\ln(1 - a)}{a}, \quad 0 < |a| < 1$$
and
$\frac{1}{32} \le \frac{(1 - 1/2)^3}{4} \le |\frac{(1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^3}{8\cdot 2^{-1}\mathrm{e}^{\mathrm{i}t}}| \le \frac{(1 + 1/2)^3}{4} = 27/32 < 1$ in (2).

Proof of the integral representation (1):
Using Cauchy integral formula, we have
$$\Big[(1 + z)^{3k + 1}\Big]^{(k)}(0) = \frac{k!}{2\pi \mathrm{i}}
\oint\limits_{|z| = 1/2} \frac{(1 + z)^{3k + 1}}{z^{k + 1}} \mathrm{d} z$$
which results in (with the substitution $z = 2^{-1}\mathrm{e}^{\mathrm{i} t}$)
$$\binom{3k + 1}{k} = \frac{1}{2\pi \mathrm{i}} 
 \oint\limits_{|z| = 1/2} \frac{(1 + z)^{3k + 1}}{z^{k + 1}} \mathrm{d} z
 = 
\frac{1}{2\pi \mathrm{i}} 
\int_{-\pi}^\pi \frac{(1 + 2^{-1}\mathrm{e}^{\mathrm{i} t})^{3k + 1}}{(2^{-1}\mathrm{e}^{\mathrm{i} t})^{k + 1}} 2^{-1}\mathrm{e}^{\mathrm{i} t} \mathrm{i}\, \mathrm{d} t.$$
The desired result follows.
