How to show that $\lim\limits_{n\to\infty}n^{2/3}a_{n}=\sqrt[3]{2}/\Gamma{(1/3})$ Let $$\left(\dfrac{1+x}{1-x}\right)^{1/3}=\sum_{n=0}^{\infty}a_{n}x^n,|x|<1$$
Show that
$$\lim_{n\to\infty}n^{2/3}a_{n}=\dfrac{\sqrt[3]{2}}{\Gamma{\left(\dfrac{1}{3}\right)}}$$
 A: Using the binomial theorem and the Cauchy convolution we have:
$$\begin{eqnarray*}[z^h]\left(\frac{1+z}{1-z}\right)^{1/3}&=&(-1)^h\sum_{l\leq h}\binom{-1/3}{l}\binom{1/3}{h-l}(-1)^l\\ &=& (-1)^h\binom{-1/3}{h}\cdot\phantom{}_2 F_1\left(-\frac{1}{3},-h;\frac{2}{3}-h,-1\right).\end{eqnarray*}\tag{1}$$
Since:
$$\lim_{h\to+ \infty}\phantom{}_2 F_1\left(-\frac{1}{3},-h;\frac{2}{3}-h,-1\right)=(1+1)^{1/3}=2^{1/3},\tag{2}$$
we just have to study the asymptotic behaviour of:
$$\begin{eqnarray*}\binom{-1/3}{n}&=&\frac{\Gamma(2/3)}{\Gamma(n+1)\Gamma(2/3-n)}=\frac{\Gamma(2/3)\Gamma(1/3+n)}{\Gamma(n+1)\Gamma(2/3-n)\Gamma(1/3+n)}\\
&=&\frac{\Gamma(2/3)\sin(\pi n+\pi/3)}{\pi}\cdot\frac{\Gamma(n+1/3)}{\Gamma(n+1)}\\&=&\frac{(-1)^n\,\Gamma(2/3)\sqrt{3}}{2\pi}\cdot n^{-2/3}\left(1+O\left(\frac{1}{n}\right)\right)\end{eqnarray*}\tag{3}$$
so the limit is:
$$\frac{2^{1/3}}{\Gamma(1/3)}\tag{4}$$
due to the Euler reflection formula.
A: Write
$$
\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{(1+x)^{1/3} - 2^{1/3}}{(1-x)^{1/3}} + \frac{2^{1/3}}{(1-x)^{1/3}} =: f(x)+g(x).
$$
The function $f$ is continuous on the disk $|z| \leq 1$, so by Darboux's method (Thm. VI.14 in Analytic Combinatorics) we have
$$
[x^n] f(x) = o\left(\frac{1}{n}\right).
$$
The coefficients of the second series are given by
$$
[x^n] g(x) = 2^{1/3} (-1)^n \binom{-1/3}{n} = \frac{2^{1/3}}{\Gamma(1/3) n^{2/3}} + O\left(\frac{1}{n^{5/3}}\right)
$$
as shown by Jack D'Aurizio.
Combining these yields
$$
\begin{align}
[x^n]\left(\dfrac{1+x}{1-x}\right)^{1/3} &= [x^n] f(x) + [x^n] g(x) \\
&= \frac{2^{1/3}}{\Gamma(1/3) n^{2/3}} + o\left(\frac{1}{n}\right),
\end{align}
$$
so that
$$
n^{2/3} [x^n]\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{2^{1/3}}{\Gamma(1/3)} + o\left(\frac{1}{n^{1/3}}\right).
$$
