Results 1 to 2 of 2

Thread: Region of convergence of an infinite series

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Region of convergence of an infinite series

    Find the region of convergence of the power series \sum  _{n=0}^{+\infty} a^{n^2} z^n where a \in \mathbb{C}.
    According to my class notes, if I have a power series of the form \sum _{n=0}^{+\infty} a_n z^n, the radius of convergence is equal to R=\lim _{n \to \infty}   \frac{|a_n|}{|a_{n+1}|}. So I applied the formula and I got that R= \lim _{n \to \infty}  \frac{1}{|a^{2n+1}|}.
    I wrote a=re^{i \theta} \Rightarrow a^{2n+1}=r^{(2n+1)}  e^{i(2n+1)\theta}. So I think that \lim _{n \to \infty}  |a^{2n+1}| depends on |a|=r. If |a|<1 the series converges while if |a|>1 the series diverges, \forall z \in \mathbb{C}. So the radius of convergence would be 1, although it seems totally wrong to me since I think I should have had a condition on z, not only on a. For instance if a=1 I fall in the case when I have \sum _{n=0}^{+\infty} z^n and the radius of convergence is indeed 1 but because of a condition on z.
    So I'd like to know what I'm doing wrong.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    Find the region of convergence of the power series \sum  _{n=0}^{+\infty} a^{n^2} z^n where a \in \mathbb{C}.
    According to my class notes, if I have a power series of the form \sum _{n=0}^{+\infty} a_n z^n, the radius of convergence is equal to R=\lim _{n \to \infty}   \frac{|a_n|}{|a_{n+1}|}. So I applied the formula and I got that R= \lim _{n \to \infty}  \frac{1}{|a^{2n+1}|}.
    I wrote a=re^{i \theta} \Rightarrow a^{2n+1}=r^{(2n+1)}  e^{i(2n+1)\theta}. So I think that \lim _{n \to \infty}  |a^{2n+1}| depends on |a|=r. If |a|<1 the series converges while if |a|>1 the series diverges, \forall z \in \mathbb{C}. So the radius of convergence would be 1, although it seems totally wrong to me since I think I should have had a condition on z, not only on a. For instance if a=1 I fall in the case when I have \sum _{n=0}^{+\infty} z^n and the radius of convergence is indeed 1 but because of a condition on z.
    So I'd like to know what I'm doing wrong.
    The formula R= \lim _{n \to \infty}  \frac{1}{|a^{2n+1}|} is correct. If |a| > 1 then this limit is 0. In other words, the radius of convergence is then 0 (which means that the series does not converge for any value of z except 0). Similarly, if |a| < 1 then the limit is infinite. In other words, the radius of convergence is then infinity (which means that the series converges for all z). Finally, if |a| = 1 then you are correct to say that the radius of convergence is 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. convergence of infinite series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 15th 2010, 06:47 PM
  2. Replies: 7
    Last Post: Oct 12th 2009, 10:10 AM
  3. convergence of an infinite series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 17th 2008, 04:54 AM
  4. Infinite series : convergence.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Jul 11th 2008, 02:09 PM
  5. Convergence of infinite series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 12th 2008, 02:11 PM

Search tags for this page

Click on a term to search for related topics.

/mathhelpforum @mathhelpforum