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Sequence Prove

  1. May 1, 2012 #1
    If( an) convergent sequence,prove that lim n goes to infinity an = lim n goes to infinity a2n+1.


    I think a2n+1 is subsequence of (an ) and for this reason their limit is equal.

    but ı don't know where and how to start..
     
  2. jcsd
  3. May 1, 2012 #2

    sharks

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    Your wording isn't very clear to me and doesn't make much sense. Is that exactly how the question appears in your book/notes?
     
  4. May 1, 2012 #3

    sharks

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    OK, your problem appears to be:
    [tex]\lim_{n\to\infty} a_n = \lim_{n\to\infty} a_{2n+1}[/tex]
    You are already told that [itex]a_n[/itex] converges, so you have to show that [itex]\lim_{n\to\infty} a_{2n+1}[/itex] also converges to the same limit.

    Every subsequence of a convergent sequence is convergent, with the same limit. So, if [itex]a_n[/itex] converges, then [itex]a_{2n}[/itex] and [itex]a_{2n+1}[/itex] are convergent as well.

    Let [itex]\lim_{n\to\infty} a_n = L[/itex], then [itex]\lim_{n\to\infty} a_{2n} = \lim_{n\to\infty} a_{2n+1} = L[/itex]
     
    Last edited: May 1, 2012
  5. May 1, 2012 #4

    LCKurtz

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    Of course the limit of a subsequence is the same as the limit of the sequence. But it looks to me like that's what you are trying to prove, albeit in a special case. So prove it from the basics. Write down an ##\epsilon - n## definition of what it means for the original sequence to have a limit, then write down the same kind of statement for what you need to prove. Use what you are given to get what you need.
     
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