Finding $\lim_{n\rightarrow \infty} \sqrt[n]{\binom{\alpha n}{n}}$ I've seen in my book a solution to the case $\alpha=2$ by showing $a_n=\binom{2 n}{n}=\prod_{i=1}^n\frac{(2i-1)(2i)}{i^2}(=b_i)$, so $\lim \sqrt[n]{a_n}=\lim b_n=4$
for $\alpha=3$, by denoting $a_n=\binom{3n}{n}$ they show $\frac{a_{n+1}}{a_n}=\frac{(3n+1)(3n+2)(3n+2)}{(n+1)(2n+1)(2n+2)}\rightarrow \frac{27}{4}$ which is also the limit of $\sqrt[n]{a_n}$ (there was a question before that proved  $\lim_{n \rightarrow \infty} \sqrt[n]{a_n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$ prior).
Is there some generalization of this for all $\alpha\in\mathbb{N}$? 
 A: Note that by Stirling's formula, $n! \sim (\frac n e)^n\sqrt{2\pi n}$
Therefore,
$$\begin{align}
{an \choose n}&\sim \frac {(\frac{an}e)^{an}\sqrt{2\pi an}} {(\frac{n}e)^{n}\sqrt{2\pi n}(\frac{(a-1)n}e)^{(a-1)n}\sqrt{2\pi (a-1)n}}\\
&\sim (\frac a {a-1})^{an}(a-1)^n\sqrt{\frac a {2\pi n(a-1)}}
\end{align}
$$
Since $\lim_{n \to \infty}({\frac a {2\pi n(a-1)}})^{\frac 1 {2n}} =1$, $$\lim_{n \to \infty}{an \choose n}^{{\frac 1 n}}=(a-1)(\frac a {a-1})^a$$
A: Yes, there is a general result for $\alpha\in\mathbb{N}$.
We have that
$$\frac{a_{n+1}}{a_n}=
\frac{(\alpha n)^{\alpha}+o(n^{\alpha})}{(n+1)\cdot\left(((\alpha-1)n)^{\alpha-1}+o(n^{\alpha-1})\right)}
=\frac{\alpha^{\alpha}+o(1)}{(\alpha-1)^{\alpha-1}+o(1)}\to
\frac{\alpha^{\alpha}}{(\alpha-1)^{\alpha-1}}$$
where $a_n=\binom{\alpha n}{n}$.
Hence
$$\lim_{n\rightarrow \infty} \sqrt[n]{\binom{\alpha n}{n}}
=\lim_{n\rightarrow \infty} \sqrt[n]{a_n}
=\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}
=\frac{\alpha^{\alpha}}{(\alpha-1)^{\alpha-1}}.
$$
A: We have
$$
\binom{\alpha n}{n}=\frac{\Gamma\left(\alpha n+1\right)}{\Gamma\left((\alpha-1)n+1\right)n!}
$$
so applying Gautschi's Inequality, we get
$$
\begin{align}
\frac{\binom{\alpha n+\alpha}{n+1}}{\binom{\alpha n}{n}}
&=\frac{\Gamma\left(\alpha n+\alpha+1\right)}{\Gamma\left((\alpha-1)n+\alpha\right)(n+1)!}\frac{\Gamma\left((\alpha-1)n+1\right)n!}{\Gamma\left(\alpha n+1\right)}\\
&=\frac1{n+1}\frac{\Gamma\left(\alpha n+\alpha+1\right)}{\Gamma\left(\alpha n+1\right)}\frac{\Gamma\left((\alpha-1)n+1\right)}{\Gamma\left((\alpha-1)n+\alpha\right)}\\
&\sim\frac1n(\alpha n)^\alpha((\alpha-1)n)^{1-\alpha}\\
&=\frac{\alpha^\alpha}{(\alpha-1)^{\alpha-1}}
\end{align}
$$
