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Math Help - Convergence of Sequence

  1. #1
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    Convergence of Sequence

    Prove that if the sequence a(n+k) converges to l then the sequence a(n) also converges to l.

    My proof basically looks like this:
    there exists N s.t. n+k>n>N.
    Assume the sequence a(n+k) converges to l and that the sequence a(n) diverges. By the definition: there exists L>0 s.t. for every K>0, there exists N that is an integer s.t n which is also an integer n>N -> |a(n)-L|<K

    This means that if the sequence a(n) converges, then the sequence a(n+k) also converges because n+k>n>N. But if a(n) diverges, by the definition of divergence (eg not(definition of convergence)) a(n+k) also diverges. And since a(n+k) only converges when a(n) converges, there is a contradiction with my initial assumption where a(n+k) converges and a(n) diverges.

    The problem that im having is with my first line, there exists N s.t. n+k>n>N. Is this a valid statement? I keep getting the feeling that there is something wrong with it.
    Last edited by ah-bee; May 9th 2008 at 06:11 AM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    The problem that im having is with my first line, there exists N s.t. n+k>n>N. Is this a valid statement? I keep getting the feeling that there is something wrong with it.
    It is valid (if k,\,n>0) but I don't understand why you need it. Is it to say that if n>N then n+k>N ?

    Quote Originally Posted by ah-bee View Post
    Prove that if the sequence a(n+k) converges to l then the sequence a(n) also converges to l.

    My proof basically looks like this:
    there exists N s.t. n+k>n>N.
    Assume the sequence a(n+k) converges to l and that the sequence a(n) diverges. By the definition: there exists L>0 s.t. for every K>0, there exists N that is an integer s.t n which is also an integer n>N -> |a(n+k)-L|<K

    This means that if the sequence a(n) converges, then the sequence a(n+k) also converges because n+k>n>N.
    Why not using this method to prove directly that if (a_{n+k}) converges then (a_n) converges too ? I think it would make your proof clearer :

    Let's assume that (a_{n+k}) converges, that is to say that \forall \varepsilon,\,\exists N\,|\,n>N \implies |a_{n+k}-L|<\varepsilon hence by taking N_0=? we get that \forall \varepsilon,\,\exists N_0\,|\,n>N_0 \implies |a_{n}-L|<\varepsilon hence (a_n) converges.

    It's exactly the idea you used but the proof is both shorter and clearer.

    Hope that helps.
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