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Real Analysis - Radius of Convergence

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that [tex]\sum[/tex]anxn has finite radius of convergence R and that an >= 0 for all n. Show that if the series converges at R, then it also converges at -R.

    2. Relevant equations

    3. The attempt at a solution
    Since the series converges at R, then I know that [tex]\sum[/tex]anRn = M.

    At -R, the series is the following: [tex]\sum[/tex]an(-R)n = [tex]\sum[/tex](-1)nanRn.

    I'm not sure where to go from here. I thought I needed to use the alternating series test, but how can I know that a1 >= a2 >= ... >= an for all n? Do I know this because the series converges? Thanks for your help.
     
  2. jcsd
  3. Apr 29, 2008 #2

    Dick

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    Right. You can't use the alternating series test. How about a comparison test?
     
  4. Apr 29, 2008 #3
    Thanks! So since [tex]\sum[/tex]anRn converges, and an(-R)n <= anRn for all n, then [tex]\sum[/tex]an(-R)n converges.
     
  5. Apr 29, 2008 #4

    Dick

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    Sorry! That's wrong. I'm clearly asleep at the wheel. That's convergence for sequences. And this sort of argument only shows that the partial sums are bounded, not that they converge. Do you know the Dirichlet convergence test?
     
    Last edited: Apr 29, 2008
  6. Apr 29, 2008 #5
    I don't know that one. But the comparison test in my book says the following:

    Let [tex]\sum[/tex]an be a series where an >=0 for all n.
    (i) If [tex]\sum[/tex]an converges and |bn| <= an for all n, then [tex]\sum[/tex]bn converges.

    If I let an = anRn, this is >=0 for all n. And if I let bn = an(-R)n, then I have |bn| <= an for all n, so the series converges, right? What's wrong with this statement?
     
  7. Apr 29, 2008 #6

    Dick

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    Nothing wrong with that. Unfortunately, I wasn't thinking of that comparison test. Hence the panic attack. Carry on.
     
  8. Apr 29, 2008 #7
    Ok! :smile: Thanks once again for your help.
     
  9. Apr 29, 2008 #8

    HallsofIvy

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    What's wrong with using the comparison test is that it only applies to positive series. Certainly -n< (1/2)n for all n, but we can't use that to conclude that [itex]\sum -n [/itex] converges!

    The crucial point is that every an is positive. That means that [itex]\sum a_n x^n[/itex], for x negative is an alternating series. What is true of alternating series?
     
    Last edited: Apr 29, 2008
  10. Apr 29, 2008 #9
    If a1 >= a2 >= ... >= an >= ... >= 0 and lim(an) = 0, then the alternating series [tex]\sum[/tex](-1)nan converges. But, like I said before, do I know that a1 >= a2 >= ... >= an because the series converges at R?
     
  11. Apr 29, 2008 #10

    Dick

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    All of terms a_n*R^n are positive and it's convergent. The series is absolutely convergent. Nothing can go wrong here.
     
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