Complex series: Circle of convergence

  1. 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,
  2. jcsd
  3. 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.
  4. tiny-tim

    tiny-tim 26,041
    Science Advisor
    Homework Helper

    Hi Niles! :smile:
    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. :smile:
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?