$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ 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   .
 A: 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.
A: 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
$$\sum_{k=0}^\infty \frac{1}{k+1}{3k+1 \choose k}\frac{1}{2^{3k+3}}=\frac{3-\sqrt{5}}{4}.$$
A: The given sum is a specific case of the generating function of the Fuss–Catalan numbers: $$\sum_{k=0}^\infty\frac{n}{mk+n}\binom{mk+n}{k}z^n=F_m(z)^n,$$ where $w=F_m(z)$ is the solution of $w=1+zw^m$ analytic in a neighborhood of $z=0$.
Since $\frac1{k+1}\binom{3k+1}{k}=\frac2{3k+2}\binom{3k+2}{k}$, our sum is $\lambda^2$, where $\lambda=\frac12 F_3\left(\frac18\right)$ satisfies $2\lambda=1+\lambda^3$. The solutions of this equation are $\lambda=1$ and $\lambda=(-1\pm\sqrt5)/2$; the correct one $\color{blue}{\lambda=(\sqrt5-1)/2}$ can be found in a number of ways, e.g. by estimating $F_3(1/8)$ from above. A related fact here is that the radius of convergence of $F_m(z)$ equals $z_m=(m-1)^{m-1}/m^m$, with $F_m(z_m)=m/(m-1)$.
There are numerous proofs of the formula stated at the beginning. A fairly short one (which also generalizes to non-integral values of $m$ and $n$) uses the Lagrange–Bürmann formula which states that, given analytic functions $\phi$ and $\psi$ with $\phi(0)\neq0$, if we let $f(w)=w/\phi(w)$ then, using the "coefficient-of" notation, $$[z^k]\psi\big(f^{-1}(z)\big)=\frac1k[w^{k-1}]\big(\psi'(w)\phi(w)^k\big).$$ We apply it to $\phi(w)=(1+w)^m$ and $\psi(w)=(1+w)^n$.
