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Convergence of sequences proof

  1. Oct 19, 2013 #1
    Given a sequence ## <x_n> ##, let ## <x_{n+1}> ## denote the sequence whose nth term for each ## n \in \mathbb{N} ## is ## x_{n+1} ##. Show that if ## <x_n> ## converges then ## < x_{n+1} ## converges and they have the same limit.

    my attempt thus far

    given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n > N implies ## |x_n - l | < \epsilon ## where l is the limit of ## x_n##.

    now we need to prove that for given ## \epsilon > 0 ## ##\exists N_1 \in \mathbb{R} ## s.t. ##n > N_1 ## implies ## |x_{n+1} - m | < \epsilon ## where m is the limit of ## x_{n+1} ## and m = l

    heres what I have thus far

    consider ## |x_n - l |## :

    ## |x_n - l | = |x_n - l + x_{n+1} - x_{n+1} | = |(x_n - x_{n+1}) + (x_{n+1} - l ) | \leq |x_n - x_{n+1} | + |x_{n+1} - l| \leq |x - x_{n+1} | + |x_{n+1} - l | \leq |x_{n+1} - x | + |x_{n+1} -l | ##

    any ideas how to proceed? or if I'm even doing it correctly
     
  2. jcsd
  3. Oct 19, 2013 #2
    Not a bad try. But you want to show [itex] |x_{n+1}-l|\le \epsilon [/itex] so you should start your inequality with that instead of [itex] |x_n-l| [/itex]. Also when you break up the absolute value into two pieces, use what you know about the size of [itex] |x_n-l|[/itex] and [itex]|x_n-x_{n+1}| [/itex]. I'm not sure why you introduced "x", what is it?
     
  4. Oct 19, 2013 #3

    HallsofIvy

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    If [itex]\{x_n\}[/itex] converges to L, then, given any [itex]\epsilon>0[/itex], there exist N such that if n> N then [itex]|x_n- L|< \epsilon[/itex]. Here, your second sequence, lets call it [itex]\{a_m\}[/itex] has the property that [itex]a_m= x_{n+1}[/itex]. Do you see that, for the same [itex]\epsilon[/itex], we can our new "N" to be the old N plus 1?

    (This is a different method than ArcanaNoir's. Both will work.)
     
  5. Oct 19, 2013 #4
    The x is the first term of the sequence

    if I do x_n - x_(n+1) = x_1, x_2, x_3 ... x_n - (x_2, x_3, x_4... x_(n+1))

    how do I evaluate [itex]|x_n-x_{n+1}| [/itex]? I have no idea of it's size
     
  6. Oct 19, 2013 #5
    No, I don't see that - I mean it kind of makes sense, but I'm not all that convinced.

    Thank you for your help
     
  7. Oct 19, 2013 #6
    Have you learned about Cauchy sequences?

    At any rate, you know [itex] |x_n-l|<\epsilon [/itex] for all n>N. Well if n>N, isn't n+1 > N?
     
  8. Oct 19, 2013 #7
    I have not learned about Cauchy sequences, all we have done is the definition of convergence and this is a question on my problem sheet.


    I agree with the highlighted part, but I do not see how it helps.

    thanks,
     
  9. Oct 19, 2013 #8

    Dick

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    Maybe I just don't get what you don't get, but if

    doesn't that mean [itex] |x_{n+1}-l|<\epsilon [/itex]? It's really simpler than what you think.
     
    Last edited: Oct 19, 2013
  10. Oct 20, 2013 #9
    Yeah that makes sense actually,

    but for a different proof:

    I got from here ## |x_{n+1} - l| \leq |x_{n+1} - x_n| + |x_n - l | < |x_{n+1} - x_n| + \epsilon ## how do I evaluate ## |x_{n+1} - x_n|##?
     
  11. Oct 20, 2013 #10

    Dick

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    That's really not a very good way to go. Because about the only thing you can say about ## |x_{n+1} - x_n|## is that since ##|x_n-l|<\epsilon## AND ## |x_{n+1} - l| < \epsilon## then ## |x_{n+1} - x_n| < 2\epsilon##, but you need the bound on ## |x_{n+1} - l|## to say that.
     
  12. Oct 20, 2013 #11
    hm I see, but I just cannot seem to construct a proof using that if x_n is bounded then there n>N for some N then n+1 > N also
     
  13. Oct 20, 2013 #12

    Dick

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    I don't understand why you don't see it. If n>N then n+1>N. It's automatic. The same N that works for the sequence ##a_n## works for the sequence ##b_n## where ##b_n=a_{n+1}##.
     
  14. Oct 20, 2013 #13
    So that's all I'd have to state for the proof?
     
  15. Oct 20, 2013 #14

    Dick

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    Yes, a proof could be that simple. Write it in a way that you understand it.
     
  16. Oct 20, 2013 #15
    Thank you very much Dick, and everyone else who helped :)!
     
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