Please more help me to find the convergence interval and the sum -by using residue theory- of the series. The sum is that $$\sum_{n=0}^\infty \binom{3n}{2n} x^n$$ 
First of all, I need to check whether the sum converges or not and if it is convergent, which points?
I am using ratio test. 
$$ \vert\frac{a_{n+1}(x)}{a_n (x)}\vert= \frac{(3n+3)!}{(2n+2)!(n+1)!} \frac{(2n)!n!}{(3n)!}\vert x\vert =\frac{27n^3+ \dots}{4n^3+\dots}\vert x\vert \to \frac{27}{4}\vert x\vert $$ as $n \to \infty$
Thus, the series converges absolutely for $\vert x\vert \lt 4/27$
Now I Will check the convergence of the series for $x=4/27$ 
For that, I am using Raabe's test 
$$\lim_{n \to \infty } n(1-\frac{a_{n+1}(4/27)}{a_n(4/27)}) =\lim_{n\to \infty}n(1-\frac{27n^3+54n^2+33n+6}{4n^3+12n^2+8n+2}\frac{4}{27})\gt 1 $$ 
So the series converges for $x=4/27$ 
Now, let's check that the series converges or not for $x=-4/27$ 
In this case, $$a_n(-4/27)=(-1)^na_n(4/27)$$
$$\frac{a_{n+1}(4/27)}{a_n(4/27)}=\frac{108n^3+216n^2+132n+24}{108n^3+324n^2+216n+54}\lt 1$$
So, $a_{n+1}(4/27)\lt a_n(4/27), n=0,1,\dots$ Thus, $a_n(4/27)$is strictly decreasing as $n\to \infty$
To show that $a_n(4/27)\to 0$ as $n\to \infty$ I need to use stirling's theorem. 
But I dont know how to apply the theorem. Please, can you complete this part by demostrating clearly? By the way, hopefully, so far everything is correct. If you see any mistake, please can you notify/inform me? Thank you. 
Next, I want to find the sum
$$\sum_{n=0}^{\infty}\binom{3n}{2n}x^n= \frac {1}{2\pi i}\sum_{n=0}{\infty}\int_{C_R}\frac{(z+1)^{3n}}{z^{2n+1}}x^n dz= \frac{1}{2\pi i}\int_{C_R}\sum_{n=0}^{\infty}[\frac{(z+1)^3x}{z^2}]^n \frac{dz}{z}=$$ 
$$\frac{1}{2\pi i}\int_{C_R}\frac{1}{1-\frac{(z+1)^3x}{z^2}}\frac{dz}{z}=\frac{1}{2\pi i}\int_{C_R}\frac{z}{z^2-(z+1)^3x}dz$$
The Cheney of infinite sum and integration is justified when the convergence of the infinite sum is uniform on the given circle. I need to check that for$|z|=2$ 
$$\vert \frac {(z+1)^3x}{z}\vert \le 27/4|x|<1 $$ when $|x|\lt 4/27$
So, I can take $R=2$ and $|x|<4/27$ 
Now, let's evaluate the integral by applying residue theory
What is the roots of $z^2-(z+1)^3x=0$ ?
But ıcannot calculate, and then I dont know how to decide that I need to find which points' residues. And after here,  I dont know how to find the sum. -I guess I need to find numerical result but how? And what?-
Again hopefully so far every thing is correct. Please check the second part and help me to complete final part. 
I am trying to solve such questions by myself at first time. I showed all what I know as you see. I want to learn perfectly  thus, if  like mine, you help me to solve and continue my solution in detail and step by step, I am grateful of you. Thank you so much. :))
 A: For small $|x|$ your equation has two roots inside the circle (note that at $x=0$ it is $z^2=0$, which has a double root at $x=0$).  It may be more convenient to use $(1+1/w)^{3n}w^{2n-1}$ instead of $(1+z)^{3n} z^{-2n-1}$, so you get only one root inside the circle.  You can then express your answer in terms of that root.  There are formulas for solving cubics, but they are rather messy.
FWIW, Maple gives the answer as $$ \dfrac{2 \cos\left(\frac{1}{3} \arcsin(3 \sqrt{3x}/2)\right)}{\sqrt{4-27x}}$$
A: To apply Stirling's approximation to your term ${3n \choose 2n}=\frac {(3n)!}{(2n)!n!}$ we plug it in.  It says $n! \sim \frac {n^n}{e^n} \sqrt {2\pi n}$, so ${3n \choose 2n}\sim \frac {(3n)^{3n}e^ne^{2n}}{(2n)^{2n}n^ne^{3n}}\sqrt{\frac {3n}{2 \pi n(2n)}}=\frac {3^{3n}}{2^{2n}}\sqrt{\frac 3{4\pi n}}=(\frac 32)^{2n}3^n\sqrt{\frac 3{4\pi n}}$
A: If $a_n>0$ for all $n$ and $\sum_{n=0}^\infty a_n x^n$ converges for $x=r>0$ then it also converges for $x=-r$ (and in fact for all $z\in\mathbb C$ with $|z|\le r$).
