Using Cauchy Hadamard to prove the radius of convergence for Binomial Series The general binomial series is:
$$(1 + x)^\alpha =  1 + \alpha x + \dfrac {\alpha (\alpha - 1) } {2!} x^2 + \dfrac {\alpha (\alpha - 1) (\alpha - 2) } {3!} x^3 + \cdots$$
for a real $\alpha$ and $|x| < 1$
and it is known that $R = 1$ is its radius of convergence.
Proofs are easily findable online using the ratio test. But I am struggling to use the Cauchy Hadamard theorem on this.
I want to show that $$\left|\frac {\prod \limits_{k \mathop = 0}^{n - 1}(\alpha - k) } {n!}\right|^{\frac 1 n}$$ tends to $1$.
The problem in my view is that if $\alpha$ is positive, the $\frac 1 {n!}$ dominates and then the limit tends to $0$. And then I'm not sure how to handle the $\alpha$ nonpositive. How should I proceed?
 A: Denote $a_n = \left|\frac{\prod_{k=0}^{n-1}(\alpha-k)}{n!}\right|$.
$$\frac{a_{n+1}}{a_n} = \frac{|\alpha-n|}{n+1}\to 1$$
So we also have $a_n^{\frac{1}{n}}\to 1$.
A: It is possible to prove that the limit is $1$ directly. Note that
we need to assume that $\alpha\notin\{0,1,2,\ldots\}$, otherwise
the limit is zero.
Suppose that $\alpha\notin\{0,1,2,\ldots\}$. Let $\varepsilon\in(0,1)$
be arbitrary. Note that $\left|\frac{\alpha-k}{k+1}\right|\rightarrow1$
as $k\rightarrow\infty$, so we can choose $N$ such that $\left|\frac{\alpha-k}{k+1}\right|\in(1-\varepsilon,1+\varepsilon)$
whenever $k>N$. For $n>N$, we have that
\begin{eqnarray*}
 &  & \left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}\\
 & = & \left\{ \frac{\prod_{k=0}^{N}\left|\alpha-k\right|}{(N+1)!}\right\} ^{\frac{1}{n}}\left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}.
\end{eqnarray*}
Observe that $\left\{ \frac{\prod_{k=0}^{N}\left|\alpha-k\right|}{(N+1)!}\right\} ^{\frac{1}{n}}\rightarrow1$
as $n\rightarrow\infty$ because $\frac{\prod_{k=0}^{N}\left|\alpha-k\right|}{(N+1)!}>0$.
For each $k>N$, we have $1-\varepsilon\leq\left|\frac{\alpha-k}{k+1}\right|\leq1+\varepsilon$,
so
\begin{eqnarray*}
(1-\varepsilon)^{\frac{n-N-1}{n}} & \leq & \left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}\leq(1+\varepsilon)^{\frac{n-N-1}{n}}.
\end{eqnarray*}
On one hand,
\begin{eqnarray*}
 &  & \limsup_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}\\
 & = & \lim_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{N}\left|\alpha-k\right|}{(N+1)!}\right\} ^{\frac{1}{n}}\cdot\limsup_{n\rightarrow\infty}\left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}\\
 & = & \limsup_{n\rightarrow\infty}\left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}\\
 & \leq & \lim_{n\rightarrow\infty}(1+\varepsilon)^{\frac{n-N-1}{n}}\\
 & = & 1+\varepsilon.
\end{eqnarray*}
That is, $\limsup_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}\leq1+\varepsilon$.
Since $\varepsilon$ is arbitrary, so $\limsup_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}=1$.
(It is crucial that we discharge $N$ before we discharge $\varepsilon$
because $N$ depends on $\varepsilon$.)
On the other hand,
\begin{eqnarray*}
 &  & \liminf_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}\\
 & = & \lim_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{N}\left|\alpha-k\right|}{(N+1)!}\right\} ^{\frac{1}{n}}\cdot\liminf_{n\rightarrow\infty}\left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}\\
 & = & \liminf_{n\rightarrow\infty}\left\{ \prod_{k=N+1}^{n-1}\frac{|\alpha-k|}{k+1}\right\} ^{\frac{1}{n}}\\
 & \geq & \lim_{n\rightarrow\infty}(1-\varepsilon)^{\frac{n-N-1}{n}}\\
 & = & 1-\varepsilon.
\end{eqnarray*}
Since $\varepsilon$ is arbitrary, we have $\liminf_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}=1$.
Finally, $\limsup_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}=\liminf_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}=1$,
so $\lim_{n\rightarrow\infty}\left\{ \frac{\prod_{k=0}^{n-1}\left|\alpha-k\right|}{n!}\right\} ^{\frac{1}{n}}=1$.
Remark:
In the above, we have used the fact that: If $(a_{n})$, $(b_{n})$
are bounded sequence of non-negative real numbers and $a_{n}\rightarrow a$,
then $\limsup_{n}(a_{n}b_{n})=a\cdot\limsup_{n\rightarrow\infty}b_{n}$.
This fact can be proved as follow: If $a=0$, then $RHS=0$. $LHS=0$
because $a_{n}\rightarrow0$ while $(b_{n})$ is bounded. Suppose
that $a\neq0$. Recall that limsup is the largest cluster point of
a sequence, so we can choose a subsequence $(a_{n_{k}}b_{n_{k}})$
such that $\limsup_{n\rightarrow\infty}(a_{n}b_{n})=\lim_{k\rightarrow\infty}(a_{n_{k}}b_{n_{k}})$.
Note that $b_{n_{k}}=(a_{n_{k}}b_{n_{k}})/a_{n_{k}}$, so $\lim_{k\rightarrow\infty}b_{n_{k}}$
converges too. Now, $\limsup_{n\rightarrow\infty}(a_{n}b_{n})=\lim_{k\rightarrow\infty}(a_{n_{k}}b_{n_{k}})=\lim_{k\rightarrow\infty}a_{n_{k}}\cdot\lim_{k\rightarrow\infty}b_{n_{k}}\leq a\cdot\limsup_{n\rightarrow\infty}b_{n}$.
To prove the reverse inequality, we choose a subsequence $(b_{n'_{k}})$
such that $\limsup_{n\rightarrow\infty}b_{n}=\lim_{k\rightarrow\infty}b_{n'_{k}}$.
We have that $\limsup_{n\rightarrow\infty}(a_{n}b_{n})\geq\lim_{k\rightarrow\infty}(a_{n'_{k}}b_{n'_{k}})=\lim_{k\rightarrow\infty}a_{n'_{k}}\cdot\lim_{k\rightarrow}b_{n'_{k}}=a\cdot\limsup_{n\rightarrow\infty}b_{n}$.
