TITLE: $\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$?
QUESTION [9 upvotes]: 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   .

REPLY [3 votes]: 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 $\psi(w)=(1+w)^n$.