What is the Radius of Convergence for a Series with a Real Non-Zero Alpha?

[aruwin]

Hello.

How do I find the radius of convergence for this problem?
$\alpha$ is a real number that is not 0.

$$f(z)=1+\sum_{n=1}^{\infty}\alpha(\alpha-1)...(\alpha-n+1)\frac{z^n}{n!}$$

 

[Prove It]

Hello.

How do I find the radius of convergence for this problem?
$\alpha$ is a real number that is not 0.

$$f(z)=1+\sum_{n=1}^{\infty}\alpha(\alpha-1)...(\alpha-n+1)\frac{z^n}{n!}$$

Have you tried using the Ratio Test?

 

[aruwin]

Have you tried using the Ratio Test?

Before that, does the value of $\alpha$ matter? I mean, no matter if it's positive or negative, would it affect the radius of convergence?
I understand that we can use the ratio test to find R. And by using ratio test, I got R=1. But in the answer, it also says that if α is a positive real number, then this series terminates.
==> The radius of convergence is ∞.

 

[aruwin]

Have you tried using the Ratio Test?

If α is a positive integer, the ratio test doesn't apply because the series isn't infinite.

Consider the example α = 4. The coefficient of z^5 would be

4(4 - 1)(4 - 2)(4 - 3)(4 - 4)/5! = 0

Every coefficient after that would also be zero. So, does that mean the ratio test can only be used when α is a NEGATIVE INTEGER? What happens to the radius of convergence when alpha is a positive integer?

By the way, the ratio test gives me R= 1.