Results 1 to 2 of 2

Math Help - 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
    7
    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: April 15th 2010, 06:47 PM
  2. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  3. convergence of an infinite series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2008, 04:54 AM
  4. Infinite series : convergence.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 11th 2008, 02:09 PM
  5. Convergence of infinite series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 12th 2008, 02:11 PM

/mathhelpforum @mathhelpforum