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Complex series: Circle of convergence 
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#1
May3009, 10:06 AM

P: 1,863

1. The problem statement, all variables and given/known data
Hi all. Lets say I have a complex power series given by [tex] \sum_{n=0}^\infty c_nz^n, [/tex] where z is a complex number and c is a complex constant. Inside its circle of convergence, I can differentiate it leading to [tex] \sum_{n=0}^\infty c_nnz^{n1} = \sum_{n=0}^\infty (n+1)c_{n+1}nz^{n}. [/tex] If I want to find the circle of convergence for this series, then I can use the ratio test [tex] \frac{1}{R} = \mathop {\lim }\limits_{n \to \infty } \left {\frac{{a_{n + 1} }}{{a_n }}} \right. [/tex] 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 [tex] \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. [/tex] 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
May3009, 11:50 AM

P: 1,104

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.



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