How find the sum $\sum_{n=1}^{\infty}\frac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}$ today,I see a amazing math problem:
show that

$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}=\dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3}$$

This problem is from here. 
But I consider sometimes,and I think it maybe use Taylor therom

$$\arcsin{x}=\sum_{n=0}^{\infty}\dfrac{(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1}$$

Thank you,Now is 24:00 in beijing time,so I must go bed.I hope someone can help.Thank you 

The transformed series in Marin's post seems suspiciously like Taylor's expansion of Bring radical
$$\begin{aligned}\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} &= \sum_{n=0}^{\infty} \binom{4n}{n} \frac{1}{(3n + 1)2^{4n + 1}} \\ &= \frac{16^{1/3}}{2} \sum_{n=0}^{\infty} \binom{4n}{n} \frac{\left ( 16^{-1/3} \right )^{3n + 1}}{(3n + 1)} \\ &= \frac{1}{2^{1/3}}\sum_{n = 0}^{\infty} \frac{\left(16^{-1/3}\right)^n}{n!} \left [\frac{d^{n-1}}{dz^{n-1}} \left ( 1 + z \right )^{4n} \right ]_{z = 0}\end{aligned}$$
The last sum is the Lagrange-Burmann inversion formula applied to $\phi(z)=(1+z)^4$. It is straightforward now that the series (excluding the constant factor $1/2^{1/3}$) is one of the two real roots of the quartic $z^4 + z + 16^{-1/3}$. One can exclude $-1/2^{1/3}$ via invoking a numerical check, and thus
$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} = \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \blacksquare$$
