Can the radius of convergence be equal to $1$?

Question

Suppose $a \in \mathbb{R}$

Can the function $ \ (1+x)^a=\sum_{n=0}^{\infty} \binom{a}{n} x^n=\sum_{n=0}^{\infty} \frac{a(a-1)(a-2) \cdots (a-n+1)}{n!}x^n$ have radius of convergence $ \ 1$ ?

Answer: Let $R$ be the radius of convergence, then by ratio test,

$R=\lim_{n \to \infty} \frac{b_{n+1}}{b_n} $ , where $ b_n=\frac{a(a-1)(a-2) \cdots (a-n+1)}{n!}$.

When will be $ \ R=1$ ?

Help me

Answer 1

Notice that\begin{align}\frac{b_{n+1}}{b_n}&=\frac{\frac{a(a-1)\ldots(a-n)}{(n+1)!}}{\frac{a(a-1)\ldots(a-n+1)}{n!}}\\&=\frac{a-n}{n+1}\end{align}and that therefore$$\lim_{n\to\infty}\left\lvert\frac{b_{n+1}}{b_n}\right\rvert=1$$and so $R=1$. Of course, this assumes that $a\notin\mathbb{Z}^+$. Otherwise, $b_n=0$ if $n\gg0$.

Answer 2

First, we define the falling factorial as $$(a)_n=\prod_{k=0}^{n}(a-k+1)=n!\cdot{a\choose n}$$ Thus, $$b_n=\frac{(a)_n}{n!}$$ Which gives $$r_n=\frac{b_{n+1}}{b_n}\\r_n=\frac{\frac{(a)_{n+1}}{(n+1)!}}{\frac{(a)_n}{n!}}\\r_n=\frac{a-n}{n+1}$$ And thus, $$R=\lim_{n\to\infty}|r_n|=1$$