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Homework Help: Hard sequence queston (calculus)

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    if {An} converges and lim {An}=L L[tex]\neq[/tex] 0
    and {Bn} diverges then does {An X Nn} diverge? prove formally


    2. Relevant equations



    3. The attempt at a solution
    IMG2.jpg

    can anyone give me a hint or show a solution?
     
  2. jcsd
  3. Sep 28, 2010 #2

    hunt_mat

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    Examine the sequences:
    [tex]
    a_{n}=1-\frac{1}{\sqrt{n}},\quad b_{n}=n
    [/tex]
     
  4. Sep 28, 2010 #3
    i dont see how that helps
     
  5. Sep 28, 2010 #4

    Dick

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    You'll probably find it easier to organize your proof as a proof by contradiction. Assume {a_n*b_n} converges, say to M, and a_n converges to a nonzero L. Can you show that means b_n would converge to M/L?
     
  6. Sep 28, 2010 #5
    yeah i thought of proof by contradiction to day, but i end up getting (M/(L+[tex]\epsilon[/tex]1)) < Bn< (m+[tex]\epsilon[/tex]2)/L

    and how do you get it converges to M/L? do i need to let epsilon = something?
     
    Last edited: Sep 28, 2010
  7. Sep 28, 2010 #6
    L<An<L +[tex]\epsilon[/tex] [tex]\leftrightarrow[/tex] [tex]\frac{1}{L +epsilon}[/tex] <[tex]\frac{1}{An}[/tex] < [tex]\frac{1}{L}[/tex]

    and M<AnBn<M+ε

    then i get [tex]\frac{M}{L+epsilon}[/tex] <Bn< [tex]\frac{M+epsilon}{L}[/tex]

    now what do i do?
     
    Last edited: Sep 28, 2010
  8. Sep 28, 2010 #7

    Dick

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    You could stop being so sloppy for one thing. What you actually want to say is for every e>0 there is an N such that for all n>N, L-e<=An<=L+e. Now if you pick e small enough so that L-e>0 (assuming L>0) then you can say 1/(L+e)<=1/An<=1/(L-e). Same sort of thing for An*Bn. Finally you want to show that you can make |Bn-M/L| as small as you want by picking e small enough. You could skip all of this epsilon-delta business if you have a theorem that says if an->A and bn->B and B is nonzero then an/bn->A/B.
     
  9. Sep 28, 2010 #8
    wait how do you get to |Bn-M/L|<e?
     
  10. Sep 28, 2010 #9

    Dick

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    You write down an expression like ...<Bn<... and subtract M/L. Then take the absolute value and show you can make the upper and lower bounds as small as you want. Are you sure you don't have a theorem about the quotient of limits?
     
  11. Sep 28, 2010 #10
    so some thing like this?
    and for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i cant use that


    IMG_0002.jpg
     
    Last edited: Sep 28, 2010
  12. Sep 28, 2010 #11

    Dick

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    No. For one thing |an-L|<e becomes L-e<an<L+e. It doesn't imply L<an. And choosing your epsilons to be L and M doesn't help at all. One more time with the question, you don't have a theorem about lim an/bn when lim bn is not zero?
     
  13. Sep 28, 2010 #12
    for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i cant use that. and i dont understand how to get the upper and lower bounds, can you please explain it?
     
  14. Sep 28, 2010 #13

    Dick

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    |An-L|<e means -e<An-L<e. And if you have the limit of quotients theorem, apply it to an*bn/an. The numerator and denominator of that both converge if you are assuming an*bn->M. Applying the theorem will save you from all this epsilon monkey business, which you don't seem to have a very good handle on. No offense.
     
  15. Sep 28, 2010 #14
    then is this right ?
    and that says DNE not one in the picture

    IMG4.jpg
     
  16. Sep 28, 2010 #15

    Dick

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    That's basically it. You should maybe mention you can apply the theorem because lim (an)=L and L is not zero.
     
  17. Sep 28, 2010 #16
    ok thanks , just out of curiosity how would you prove it with epsilons
     
  18. Sep 28, 2010 #17
    is this right using epsilons?

    Untitled.jpg
     
  19. Sep 28, 2010 #18
    Use reductio ad absurdum.
     
  20. Sep 28, 2010 #19
    please explain how you would do the contradiction
     
  21. Sep 28, 2010 #20
    1. If [itex]\lim_{n \rightarrow \infty}{\{a_{n}\}} = L, \; L \neq 0[/itex] show that, starting from some [itex]n_{0}[/itex], [itex]|a_{n}| > 0, \forall n \ge n_{0}[/itex];

    2. Assume the opposite is true, namely [itex]c_{n} \equiv a_{b} \, b_{n}[/itex] converges;

    3. Because of 1, it makes sense to define the sequence [itex]c_{n}/a_{n}, \; \forall n \ge n_{0}[/itex]. Because both [itex]\{a_{n}\}[/itex] and [itex]\{c_{n}\}[/itex] are convergent, what can we say about their quotient?
     
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