what conditions on the positive integer $ \ a \ $ imply that the radius of convergence of the power series would be $ \ 1 \ $?

24 Views Asked by At

Consider the binomial function or infinite series $ \ B_a(x)=(1+x)^a=\sum_{n=0}^{\infty} \binom{a}{n}x^n=\sum_{n=0}^{\infty} \frac{\large a(a-1)(a-2) \cdots \cdots (a-n+1)}{\large n!}x^n$.

(i) Under what conditions on the positive integer $ \ a \ $, the radius of convergence of the power series would be $ \ 1 \ $ ?

Answer:

(i)

Let $R$ be the radius of convergence.

The $n^{th}$ term of the series is $ b_n=\frac{\large a(a-1)(a-2) \cdots \cdots (a-n+1)}{\large n!}$.

Now by ratio test,

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

What would be the value of $ \ a $ from the last equation?

Is there any option so that $R$=radius of convergence$=1$ ?

Help me