# Complex series: Circle of convergence

1. May 30, 2009

### Niles

1. The problem statement, all variables and given/known data
Hi all.

Lets say I have a complex power series given by

$$\sum_{n=0}^\infty c_nz^n,$$

where z is a complex number and c is a complex constant. Inside its circle of convergence, I can differentiate it leading to

$$\sum_{n=0}^\infty c_nnz^{n-1} = \sum_{n=0}^\infty (n+1)c_{n+1}nz^{n}.$$

If I want to find the circle of convergence for this series, then I can use the ratio test

$$\frac{1}{R} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}{{a_n }}} \right|.$$

I have two questions for this:

Question #1: Does the ratio test give me the same answer regardless of I substitute e.g. n -> n+3? I.e., is it correct that

$$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 3} }}{{a_{n + 2} }}} \right|.$$

Personally, I think it does not matter, because we let n go to infinity.

Question #2: Does the ratio test only work for power series that go from n=0 to infinity, or do they also work if n start at e.g. 1 or -1?

Thank you very much in advance.

Best regards,
Niles.

2. May 30, 2009

### snipez90

Well the ratio test is typically derived from comparison to a geometric series. Moreover, one way to derive the comparison test is based on the monotone convergence theorem applied to partial sums. So I think you're right on both counts since only long term behavior matters. Especially for question 2, remember that including or excluding a finite number of terms in an infinite series does not affect convergence.

3. May 30, 2009

### tiny-tim

Hi Niles!
Your intuitive suspicions are completely correct …

limiting behaviour is only affected "near" the limit …

what happens at the other end doesn't matter!

So yes, to both #1 and #2.