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Convergence properties.

  1. Nov 17, 2012 #1
    Hello,

    I am trying to prove/disprove the following claims:

    (1) There exists a sequence a_n of positive number so that the series Ʃ a_n converges whereas the series Ʃ (a_n)^2 diverges.

    I believe I managed to disprove that. Is it indeed false?

    (2) There exists a sequence of real numbers a_n so that the series Ʃ a_n converges absolutely whereas the series Ʃ (a_n)^2 diverges.

    I proved it. Is it indeed true?

    (3) Let a_n and b_n be sequences of real numbers. Provided that lim n->∞ (b_n)/(a_n) = 1 and the series Ʃ a_n converges, then the series Ʃ b_n converges.

    I am not sure how to go about that one. Could someone please help?

    (4) Prove that if the series Ʃ a_n converges, then the series Ʃ (a_n)/(n^2) converges absolutely.

    I am not sure how to prove that. I am also not allowed to explicitly make use of the Cauchy product theorem. May you please advise how to go about it?

    Thanks in advance!
     
  2. jcsd
  3. Nov 17, 2012 #2
    The first two are the same.
     
  4. Nov 17, 2012 #3
    I agree that they are eseentially the same.
    How may I prove (3) and (4)? Any concrete suggestions?
     
  5. Nov 17, 2012 #4
    For (4) use Abel's theorem for convergence of general series in the form Ʃanbn
     
  6. Nov 17, 2012 #5
    I don't think we're allowed to use that. May you please formulate the theorem so I may be a better judge of that?
     
  7. Nov 17, 2012 #6

    micromass

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    In (2), you proved that there was an absolutely convergent series such that [itex]\sum a_n^2[/itex] diverges? May I know what your example of such a series is??

    For (3), write out what it means that the limit equals 1 and use a comparison test.
     
  8. Nov 17, 2012 #7
    Re (3), I have tried that approach but I am not allowed to express that quotient as lim b_n = lim a_n, since that would be tantamount to dividing by lim a_n (which is zero). So I am not sure how to proceed.
     
  9. Nov 17, 2012 #8

    micromass

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    You know that [itex]\lim_{n\rightarrow +\infty}\frac{b_n}{a_n}=1[/itex]. What does that mean by definition?
     
  10. Nov 17, 2012 #9
    That [|b_n - a_n|/|a_n|] < epsilon.
     
  11. Nov 17, 2012 #10

    micromass

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    No. Please give the full definition.
     
  12. Nov 17, 2012 #11
    There exists an epsilon>0 so that for every n0>=n |(b_n/a_n) - 1| < epsilon.
     
  13. Nov 17, 2012 #12

    micromass

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    No. You really should know that definition from the back of your head.
     
  14. Nov 17, 2012 #13
    Sorry, it's very late here. I meant to write: for every epsilon > 0 there exists n0 so that for every n>=n0 |(b_n - a_n) - 1| < epsilon. Better?
     
  15. Nov 17, 2012 #14
    b_n/a_n instead of b_n - a_n
     
  16. Nov 17, 2012 #15

    micromass

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    OK. So let's take [itex]\varepsilon=1[/itex]. Then there exists an [itex]n_0[/itex] so that for all [itex]n>n_0[/itex], we have

    [tex]|b_n-a_n|<|a_n|[/tex]

    Now use the comparison test.
     
  17. Nov 17, 2012 #16
    Sigma |b_n - a_n| converges (based on comparison test) and sigma a_n converges notwithstanding, does that necessarily indicate that sigma b_n converges?
     
  18. Nov 17, 2012 #17

    micromass

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    Yes. Try to figure out why.
     
  19. Nov 17, 2012 #18
    Is it simply because sigma b_n = sigma (b_n - a_n + a_n) = sum of two converging series, hence convergent itself?
     
  20. Nov 17, 2012 #19
    But you are not given that the series of an converges absolutely.
     
  21. Nov 17, 2012 #20
    Attached-see Abel's test for convergence.
     

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