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Convergence math problem

  1. Apr 8, 2009 #1
    How does one show that [itex]\sum_{n=1}^{\infty} \frac{z^n}{n}[/itex] converges [itex]\forall z s.t. |z| \leq 1, z \neq 1[/itex]

    i can show that it doesn't converge for z=1 (easy enough) but how do i do the rest of it?
  2. jcsd
  3. Apr 8, 2009 #2
    Re: Convergence

    Try ratio test and Abel's test.
  4. Apr 8, 2009 #3
    Re: Convergence

    ratio test gives [itex]\frac{z}{1+\frac{1}{n}} \rightarrow z[/itex] as n goes to infinity

    so this will be convergent for all z with |z|<1

    i dont really understand abels test - how does that help here?

    also it says you can't apply abels test on the boundary so how do i show it converges there also?
  5. Apr 8, 2009 #4
    Re: Convergence


    Abel's works on the boundary |z|=1 except at z=1. Are you allowed to use Abel's test? If not, I guess you'd have to wade through its proof? The key step is summation by parts, along with a sine trick. I'm assuming z is complex. Real case is easy.
  6. Apr 8, 2009 #5
    Re: Convergence

    ok so abels test can be used here o say that it converges on the boundary ASSUMING that we know it converges for all |z|<1 - and we know this from the ratio test.

    but abels test doesn't work on z=1 but we can tell it doesnt converge there.
    and then we are done.

    is that all acceptable?
  7. Apr 9, 2009 #6
    Re: Convergence

    There should be a theorem or proposition in your book that says:
    If [itex]\sum a_n (z-a)^n[/itex] is a given power series with radius of convergence [itex]R[/itex], then
    [tex] R = \lim_{n\to \infty} \left| \frac{a_n}{a_{n+1}} \right| [/tex]
    if this limit exists.
  8. Apr 9, 2009 #7
    Re: Convergence

    wouldn't that limit be 0 as [itex]z^n < z^{n+1} \Rightarrow \lim_{n \rightarrow \infty} |\frac{z^n}{z^{n+1}}|=0[/itex] as the n bits just go to 1 in the limit
  9. Apr 9, 2009 #8
    Re: Convergence

    Look what I posted. I didn't say [itex]z^n[/itex]. The formula I gave has [itex]a_n[/itex]'s in it.
  10. Apr 9, 2009 #9
    Re: Convergence

    Thm 1: If [tex]\left|\frac{b_{n+1}}{b_n}\right|\to L<1[/tex] then [tex]\sum b_n[/tex] converges absolutely.

    Thm 2: If [tex]\left|\frac{a_{n+1}}{a_n}\right|\to L[/tex] then [tex]R=1/L[/tex] is the radius of convergence of [tex]\sum a_n z^n[/tex].

    You can use either thm. Which one do you want to use? If you want to use Thm 1, you have to use [tex]b_n=a_n z^n[/tex]. If Thm 2, just use [tex]a_n[/tex]
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