Finding the Radius of Convergence for a Series with a Real Coefficient

[MissP.25_5]

Hello.

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

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

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 ∞.

I don't understand this part. I hope someone could explain it to me.

 

[SammyS]

Hello.

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

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

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 ∞.

I don't understand this part. I hope someone could explain it to me.

Are you sure it didn't say "if α is a positive integer, then this series terminates" ?

 

[MissP.25_5]

Are you sure it didn't say "if α is a positive integer, then this series terminates" ?

Aren't real number and integer the same thing? But yes, it is written real number. Perhaps the answer is wrong.

 

[Ray Vickson]

Aren't real number and integer the same thing? But yes, it is written real number. Perhaps the answer is wrong.

No, a real number can be positive or negative, rational or irrational. The numbers $\sqrt{2}, \pi$, etc. are all real but are not integers--in fact, they are not even rational.

 

[MissP.25_5]

No, a real number can be positive or negative, rational or irrational. The numbers $\sqrt{2}, \pi$, etc. are all real but are not integers--in fact, they are not even rational.

So, if it is written integers, what would that mean for the radius?

 

[MissP.25_5]

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?

 

[slider142]

As you have found, the series converges for every positive integer α; it is then equivalent to a polynomial in z of degree α. Since polynomials are defined for every value of z, we may say that the radius of convergence is infinite in each of these cases, since there is no finite radius beyond which the value of a polynomial is undefined.
Note that each series in this case is still infinite: it merely contains an infinite list of additive zeroes after a particular term. We are not barred from using the ratio test: it simply does not have any applicable information to give us due to the fact that the ratio is undefined as n approaches infinity.