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I Sequences, subsequences (convergent, non-convergent)

  1. May 13, 2017 #1
    Hi guys,
    I am not sure if my understanding of subsequence is right. For example I have sequence {x} from n=1 to infinity. Subsequence is when
    A) I chose for example every third term of that sequence so from sequence 1,2,3,4,5,6,7,... I choose subsequence 1,4,7...?
    B) Or subsequence is when I chose for example only first five terms so from sequence 1,2,3,4,5,6,7... I choose subsequence 1,2,3,4,5...?
    I have this problem because {x} from n=1 to infinity is non-convergent sequence and I want to know if subsequence of non-convergent sequence is also always non-convergent like in A) or we can make convergent subsequence from non-convergent sequence B).
    BY THE WAY SEQUENCE 1,2,3,4,5 IS CONVERGENT, RIGHT? :D
     
  2. jcsd
  3. May 13, 2017 #2

    mfb

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    Subsequences should be infinite. Choosing every third term is fine, just choosing the first 5 does not work.
    What do you want to show? The existence of at least one non-convergent subsequence, or that all subsequences do not converge?

    Convergence is a meaningless concept for finite sets (they are not sequences).
     
  4. May 13, 2017 #3
    I probably do not understand what is subsequence. Can you explain it in words and examples, please (not in formulas)?
     
  5. May 13, 2017 #4
    And I want to show if non-convergent sequence has always non-convergent subsequence.
     
  6. May 13, 2017 #5

    mfb

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    What is unclear? A subsequence is a sequence where you can skip elements of the parent sequence, but you have to keep infinitely many elements. You can take every third element, or every prime numbered element, or every power of 2, or whatever you want, but it has to be a sequence - it cannot end somewhere.
     
  7. May 13, 2017 #6

    Mark44

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    Here is an example: ##\{a_n\}_{n=1}^\infty## defined by ##a_n = (-1)^{n+1}## = {1, -1, 1, -1, ...}
    This sequence is divergent, but it has two convergent subsequences. One convergent subsequence is ##\{a_0, a_2, a_4, \dots, a_{2n}, \dots\}##. All elements of this subsequence are 1, so this subsequence is obviously convergent.
    The sequence of my example has another convergent subsequence, which I leave to you to find.
     
  8. May 13, 2017 #7
    Thanks guys, it is clear for me now.
    Mark44: All elements of second convergent subsequence are -1. Thanks for good example.
     
  9. May 13, 2017 #8
    I have one more question. I am curious, where in practice are subsequences used, and how can I formally write that elements of my subsequence are for example every third element of my sequence?
     
  10. May 13, 2017 #9

    mfb

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    If your original sequence has elements an, then the subsequence "every third" has elements sn=a3n. As examples, s1=a3 and s2=a6.
    It is unclear what that means. A specific subsequence either converges or not, there is nothing the "always" could apply to.
     
  11. May 13, 2017 #10
    I meant if there exists some sequence which is non-convergent and we can make such subsequence from it which would be convergent. But from our discussion and my knowledge it is impossible to find such sequence, or am I wrong?
     
  12. May 13, 2017 #11

    mfb

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    If you have a divergent sequence, there are always divergent subsequences (and you should be able to find examples easily). There can be convergent subsequences (Mark44 gave an example in post 6), but there don't have to be, it depends on the sequence.

    If you have a convergent sequence, all subsequences are convergent.
     
  13. May 13, 2017 #12

    WWGD

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    An additional condition here is that subsequences { ##a_{j_k} ##} ##\subset ## {## a_j ##} must preserve the order in the original, i.e., if for indices n,m if ## n<m ## in the original , then ##j_n < j_m##. , e.g., you cannot swap the order of the 300th, 500th terms of the original for a subsequence in the subsequence..
     
  14. May 13, 2017 #13

    WWGD

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    As in this case, to prove continuity in certain spaces, where the condition ## (a_k \rightarrow a ) \rightarrow (f(a_k) \rightarrow f(a) )## is equivalent to f being continuous.(Continuity implies Sequential Continuity but not viceversa)
     
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