How to find the limit superior of the nth root of the generalized binomial coefficient? My book says that
$$
\varlimsup_{n \to \infty} \sqrt[n]{\left|{\binom{\alpha}{n}} \right|} = 1
$$
where $\alpha \in \mathbb{R} \setminus \mathbb{N} \cup \{ 0 \}  $ and $n \in \mathbb{N}$. 
Where $\binom{\alpha}{n}$ is the generalized binomial coefficient
$$
\binom{\alpha}{n}= \frac{\alpha(\alpha-1)(\alpha - 2)\cdots(\alpha-(n-1))}{n!}
$$
How would one show this? How to think about infinite products like this one?
 A: Consider $a_n = \ln (\sqrt[n]{\left|{\binom{\alpha}{n}} \right|}) = \frac{1}{n} \ln(\left|{\binom{\alpha}{n}} \right|)$.
By Stolz-Cesaro theorem, we have for $\alpha \in \mathbb{R} \setminus \mathbb{N} \cup \{ 0 \}  $
\begin{align}
\lim_{n\to + \infty} a_n &= \lim_{n\to +\infty}\dfrac{\ln(\left|{\binom{\alpha}{n+1}} \right|) - \ln(\left|{\binom{\alpha}{n}} \right|)}{n+1 - n}\\
& = \lim_{n\to +\infty} \ln(\left|{\binom{\alpha}{n+1}} / {\binom{\alpha}{n}} \right|)\\
&=  \lim_{n\to +\infty} \ln(\frac{|\alpha - n|}{n+1}) = 0
\end{align}
Thus the limit looked for is equal to $\lim_{n\to +\infty} e^{a_n} = e^{\lim_{n\to \infty}a_n} = e^{0} = 1$, by the continuity of exponential function at $0$
Another solution: if we know the convergence radius of $(1+x)^\alpha = \sum_{n=0}^{+\infty} \binom{\alpha}{n}x^n$ is equal to 1 for $\alpha \in \mathbb{R} \setminus \mathbb{N} \cup \{ 0 \}  $, then the result is direct, since its convergence radius is also $$\dfrac{1}{\limsup_{n\to +\infty} \sqrt[n]{\left|{\binom{\alpha}{n}} \right|}) }$$. 
