(adsbygoogle = window.adsbygoogle || []).push({}); 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]

wherezis a complex number andcis a complex constant. Inside its circle of convergence, I can differentiate it leading to

[tex]

\sum_{n=0}^\infty c_nnz^{n-1} = \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 letngo to infinity.

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

Thank you very much in advance.

Best regards,

Niles.

**Physics Forums - The Fusion of Science and Community**

# Complex series: Circle of convergence

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Complex series: Circle of convergence

Loading...

**Physics Forums - The Fusion of Science and Community**