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

@JoséCarlosSantos, How to show that under certain conditions on $ \ a $ , the radius of convergence $R$ will be equal to $ \ 1$ ? — MAS, Oct 26 '18 at 17:03
@JoséCarlosSantos, Can you comment about the function? Do you recognize the above function as Newton's Generalized Binomial theorem? — MAS, Oct 26 '18 at 17:09

score 1 · Accepted Answer · answered Oct 26 '18 at 17:07

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 $n\gg0$.

answered Oct 26 '18 at 17:07

José Carlos Santos

427,504

Thank you very much for your nice answer. – MAS Oct 26 '18 at 17:10
1

@M.A.SARKAR If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Oct 26 '18 at 17:11
. I will surely accept it. One thing clear me, here $ a$ can be any rational number. Is it true? – MAS Oct 26 '18 at 17:13
Yes. Actually, $a$ can be any complex number. – José Carlos Santos Oct 26 '18 at 17:14
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No. First of all, $a\in\mathbb C$ does not imply that $a\in\mathbb{Z}^+$ (are you claiming that $i\in\mathbb{Z}^+$?). And all I said about $a\in\mathbb{Z}^+$ was that if this happens, then all $b_n$'s will be $0$ if $n$ is large enough, in which case your power series will actually be a polynomial and $R=+\infty$. – José Carlos Santos Oct 26 '18 at 17:18
Ok, I got it. Actually I got confused. – MAS Oct 26 '18 at 17:20

score 1 · Answer 2 · answered Oct 26 '18 at 17:11

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, $$R=\lim_{n\to\infty}|r_n|=1$$

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

2 Answers2