Radius of Convergence of Binomial Series Why the following is true?
$$\bigg|\lim_{n\rightarrow\infty}\frac{\alpha(\alpha-1)...(\alpha-n)}{(n+1)!}\frac{n!}{\alpha(\alpha-1)...(\alpha-n+1)}\bigg|=\lim_{n\rightarrow\infty}\frac{|\alpha-n|}{n+1}=1$$
This is found in this link.  
 A: The first equality holds because all other factors cancel, the second equality holds because for $n\gg \alpha$, $|\alpha-n|=n-\alpha$ and so $\frac{|\alpha-n|}{n+1}=\frac{n-\alpha}{n+1}=1-\frac{\alpha+1}{n+1}$ where the last fraction $\to 0$.
A: I think there is some typo in wiki. The binomial series formula is
$$\begin{array}{*{20}{l}}
{{{\left( {1 + x} \right)}^\alpha } = \sum\limits_{n = 0}^\infty  {{a_n}} {x^n} \quad \text{which uniformly converges when}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left| x \right| < 1}\\
{{a_n} = \left( {\begin{array}{*{20}{c}}
\alpha \\
n
\end{array}} \right) = \frac{{\prod\limits_{i = 0}^{n - 1} {\left( {\alpha  - i} \right)} }}{{n!}} = \frac{{\alpha \left( {\alpha  - 1} \right)...\left( {\alpha  - \left( {n - 1} \right)} \right)}}{{n!}}}
\end{array}$$
Now consider the following
$$\frac{{{a_{n + 1}}}}{{{a_n}}} = {a_{n + 1}}.\frac{1}{{{a_n}}} = \frac{{\alpha \left( {\alpha  - 1} \right)...\left( {\alpha  - n} \right)}}{{\left( {n + 1} \right)!}}\frac{{n!}}{{\alpha \left( {\alpha  - 1} \right)...\left( {\alpha  - \left( {n - 1} \right)} \right)}} = \frac{{\alpha  - n}}{{n + 1}}$$
I think this solves your confusion.
