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Math Help - Convergent sequences

  1. #1
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    Convergent sequences

    1) Can you find a sequnce of real nos. {sn} which has no convergent subsequence and yet for which {|sn|} converges.

    I am not able to think of such a sequence.

    2) If a subsequence of {sn} of terms with even subscripts converges to L, as well as the subsequence with odd subscripts (converges to L i.e.) then, prove that the given sequence converges to L.

    Here I intuitively know the theorem statement is sort of obvious, but how do I go about the proof?

    Any help will be appreciated.
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  2. #2
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    1) Let (s_n) be a sequence such that (\vert s_n \vert ) converges. Clearly (s_n) must have an infinite number of positive (or negative) terms. Take (s_{n_k}) such that 0 \leq s_{n_k} for all n_k and n_k < n_{k+1} then (s_{n_k}) is a subsequence of (s_n), but \vert s_n \vert converges.

    2)Use the fact that the odd and even numbers partition \mathbb{N}
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  3. #3
    Super Member Gamma's Avatar
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    more explicit for 2)

    I think for 2) more explicitly it is best to say:

    Given \epsilon >0, since the even and odd sequences converge to L, there should be an N_{odd} and an N_{even} such that for all odd n bigger than N_{odd} you have |x_n-L|<\epsilon and similarly for all even n bigger than N_{even} you have |x_n-L|<\epsilon.

    Then let N:=max\{N_{even},N_{odd}\} and you have for all n>N that |x_n-L|<\epsilon so the sequence converges to L as desired.
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  4. #4
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    For the first question I don't think it's possible. For if s_n were bounded, it would have a convergent subsequence by Bolzano-Weierstrass. Thus the sequence must be unbounded. But this too isn't possible because if |s_n| converges we have that by -|s_n|\le s_n\le |s_n| (nothing is lost by assuming all terms positive really) that s_n would converge. But s_n is unbounded and thus diverges, and so must |s_n|.
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