problem calculate $\sum_{n=0}^\infty {3n\choose n}\left({2\over27}\right)^{n}$
*firstly, I KNOW that this problem is duplicated. but I want to know what is wrong with my attempt.
my attempt
$$x=\sum_{n=0}^\infty{3n\choose n}{z^{2n+1}\over 2n+1}$$ satisfies $$x^3-x+z=0$$ within the convergence radius $|z|\leq{2\over3\sqrt{3}}$ by Lagrange Inversion theorem.
Thus $${dx\over dz}=\sum_{n=0}^\infty {3n\choose n}z^{2n}={1\over 1-3x^2}$$ and for $z={\sqrt2\over3\sqrt3}$, $x={\sqrt2\over\sqrt3}$ satisfies equation $x^3-x+z=0$.
i.e. $$\sum_{n=0}^\infty {3n\choose n}\left({2\over27}\right)^{n}={1\over 1-3x^2}=-1(???)$$
Just to turn @ThomasAndrews's observation into an answer, for the given value of $z$, $x$ may be equal to any of $\sqrt{\frac23},\,-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}},\,\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}$.To avoid a paradox we need $|x|<\frac{1}{\sqrt{3}}$, so we take the third root. In other words, the sum is$$\frac{1}{1-3\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right)^2}=\frac{\sqrt{3}+1}{2}\approx1.366.$$