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Convergence question

  1. Dec 3, 2008 #1
    prove or desprove that if An->1


    then An+1/An ->1


    ??
     
  2. jcsd
  3. Dec 3, 2008 #2

    Office_Shredder

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    Anytime the question says "prove or disprove" and you're not sure what to do, start by spending five minutes trying to generate a counterexample. If you can't do it, see if you can figure out what's stopping you (which usually leads to a proof that the statement is true)
     
  4. Dec 3, 2008 #3
    i know that if a series is converges and monotone
    then if An->1 then An+1 ->1

    so i construct a limit An+1/An
    n-> +infinity
    and say that if both members go to 1 then the whole expression goes to 1

    the problem is:
    i dont know if An is monotone
    and if this way is correct and will be regarded as a formal proof
    ??
     
  5. Dec 3, 2008 #4

    Office_Shredder

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    You don't need the monotone condition here. For all e>0, there exists N>0 such that n>N implies |An - 1| < e gives us that |An+1 - 1| < e for n>N also as n+1>n>N
     
  6. Dec 3, 2008 #5
    and after writing what you said
    i do the limit part?
     
  7. Dec 3, 2008 #6

    Office_Shredder

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  8. Dec 4, 2008 #7
  9. Dec 4, 2008 #8

    HallsofIvy

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    You have been posting questions here long enough to have learned to copy the problem correctly even if you do not wish to use Latex.

    Is this A_{n+1}/A_n or (A_n+ 1)/A_n ?
     
  10. Dec 4, 2008 #9

    Mark44

    Staff: Mentor

    Or maybe it's [tex]A_n + \frac{1}{A_n}[/tex]
     
  11. Dec 4, 2008 #10
    its A(n+1)/A(n)
     
  12. Dec 5, 2008 #11
    when i solve this type of question:
    how do i recognize that the expression doesnt converge
    and how do i disprove that the function converges
    in such case?
     
  13. Dec 5, 2008 #12
    i need to prove that (An)^n ->1

    but when i construct limit
    lim (An)^n
    n->+infinity

    i get 1^(+infinity) which says that there is no limit
    what do i do in this case in order to disprove that (An)^n->1

    ??
     
  14. Dec 5, 2008 #13

    HallsofIvy

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    Why do you need to prove that Ann goes to 1? Is this a completely different question?

    As to your original question, if An converges to 1, then, given any [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]1-\epsilon\le A_n\le 1+\epsilon[/itex]. If n> N, that's true for both An and An+1.
     
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