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
$$\lim_{n\to\infty}n^{2/3}a_{n}=\dfrac{\sqrt[3]{2}}{\Gamma{\left(\dfrac{1}{3}\right)}}$$

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
$$
n^{2/3} [x^n]\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{2^{1/3}}{\Gamma(1/3)} + o\left(\frac{1}{n^{1/3}}\right).
$$
