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Convergence proof

  1. Sep 20, 2011 #1
    Hi,

    I'm doing some homework from my analysis class. I honestly have no idea where to start. Any help would be appreciated.

    1. The problem statement, all variables and given/known data

    Let [itex]{a_n}[/itex] be a sequence that converges to 0, and let [itex]{b_n}[/itex] be a sequence. Prove that the sequence [itex]a_n b_n[/itex] converges to 0.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2011 #2
    If b is the limit of [itex]b_n[/itex] then you can make the terms of [itex]b_n[/itex] as close to b as you want by making n big enough. Also you can make a_n as close to 0 as you like by making n big enough. Now, if [itex]\epsilon > 0[/itex] you want to make the terms of [itex]a_nb_m[/itex] less than [itex]\epsilon[/itex] for large enough n. Now, using what I mentioned above, how can you find an n big enough?
     
  4. Sep 20, 2011 #3

    [itex]a_n=\frac{1}{n}[/itex], [itex]b_n=n^2[/itex]
    [itex]a_n b_n=n[/itex]

    [itex]\lim_{n\rightarrow \infty} a_n=0[/itex]
    [itex]\lim_{n\rightarrow \infty} a_n b_n \not= 0[/itex]
     
    Last edited: Sep 20, 2011
  5. Sep 20, 2011 #4
    I need the proof using epsilon. The prof. wants description of every step. its a senior level class
     
  6. Sep 20, 2011 #5
    no idea
     
  7. Sep 20, 2011 #6
    Try looking more closely at what I wrote.
     
  8. Sep 20, 2011 #7
    it shows, it does not converge
     
  9. Sep 20, 2011 #8
    Whoops! I didn't read the problem carefully. Estro is correct; it does not, in general, converge to anything, much less 0. My apologies!
     
  10. Sep 20, 2011 #9

    gb7nash

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    Homework Helper

    Did you copy the problem correctly? If you did, whoever made the problem up made an error.
     
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