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Hard infinite series converges problem (Real Analysis)

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data
    let bk>0 be real numbers such that Ʃ bk diverges. Show that the series Ʃ bk/(1+bk) diverges as well.
    both series start at k=1


    2. Relevant equations
    From the Given statements, we know 1+bk>1 and 0<bk/(1+bk)<1



    3. The attempt at a solution
    I've tried using comparison test but cannot find anything that's definitely less than bk/(1+bk). Iv'e tried the ratio test but i get stuck with a bunch of bk+1's that I cannot do anything with.


    Some help or at least a push in the right direction would be much appreciated. Thank you. I'm also kinda in a rush haha.
     
    Last edited: Dec 13, 2013
  2. jcsd
  3. Dec 13, 2013 #2

    Dick

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    You must be in a rush. If bk>0 then 1+bk>1, not 1+bk<1.
     
  4. Dec 13, 2013 #3

    jbunniii

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    Hint: try considering separately the cases ##b_k < 1## and ##b_k \geq 1##.
     
  5. Dec 14, 2013 #4
    Sorry but is there anymore guidance you could give? I keep failing at comparison and ratio test doesn't seem to work because we do not know if bk is decreasing, increasing or oscillating.
     
  6. Dec 14, 2013 #5
    I'm not an expert and might be wrong, but if you could perhaps derive from the given hint, if ##0<b_k<1##, then what happens with the other series? Does it pass the preliminary test? What happens to the other series if ##b_k \geq 1##?
     
  7. Dec 14, 2013 #6

    jbunniii

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    Well, if ##b_k < 1## then ##1 + b_k < 2##.

    And if ##b_k \geq 1## then ##2b_k \geq b_k + 1##.

    Try using these facts to get a lower bound for the terms of your series in each case.
     
  8. Dec 14, 2013 #7
    Here is my attempt...:

    case 1: bk<1 then 1+bk<2

    so bk/(1+bk) > bk/2 so and Ʃ[bk/2] diverges so by comparison Ʃ[bk/(1+bk)] diverges

    case 2: bk>=1 then 2bk>bk+1

    so Ʃ[bk/2bk] < Ʃ[bk/(1+bk)]
    -->Ʃ[1/2] < Ʃ[bk/(1+bk)]
    since Ʃ[1/2] diverges Ʃ[bk/(1+bk)] diverges by comparison


    I apologize for the bad format.
     
  9. Dec 14, 2013 #8

    jbunniii

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    OK, you have the right idea. Your solution is fine in case ##b_k < 1## for all ##k##, or in case ##b_k \geq 1## for all ##k##. But in general, it may be true that ##b_k < 1## for some ##k## and ##b_k \geq 1## for other ##k##. Your proof needs to handle this general situation as well.
     
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