closed form for the series

Question

I don't remember any method to compute the closed from for the following series. $$ \sum_{k=0}^{\infty}\binom{3k}{k} x^k .$$

I tried by putting $\binom{3k}{k}$ in Mathematica for different $k$ and asking for the generating function it deliver a complicated formula which is the following. $$ \frac{2\cos[\frac{1}{3} \sin^{-1}(\frac{\sqrt{3x}}{2})]}{\sqrt{4-27x}} $$

I was wondering if there is any simple form?

Answer 1

That's a power series about $\;x_0=0\;$ and whose sequence of coefficients is

$$a_k=\binom{3k}kx^k=\frac{(3k)!}{k!(2k)!}\implies\;\left|\frac{a_{k+1}}{a_k}\right|=\frac{(3k+3)!}{(k+1)!(2k+2)!}\cdot\frac{k!(2k)!}{(3k)!}|x|=$$

$$=\frac{(3k+1)(3k+2)(3k+3)}{(k+1)(2k+1)(2k+2)}|x|\xrightarrow[k\to\infty]{}\frac{27}{4}|x|$$

and thus the series converges for

$$\frac{27}4|x|<1\iff |x|<\frac4{27}$$

Answer 2

Hint: A closed form can be found by means of the Lagrange Inversion Formula.

An answer based upon this method is given at this MSE link.