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Proving a sequence is convergent.

  1. Oct 5, 2012 #1
    1. The problem is if an is convergent then prove or disprove by giving a counter example that an2 is also convergent.



    2. Since an is convergent then for all ε>0 there exists n0[itex]\in[/itex] [itex]N[/itex] such that lan-Ll<ε for all n>=n0

    So I then tried squaring (an-L) which gives an2 -2anL +L22
    How do I manipulate this to show that an2 has a limit L too?

    Or should I be looking for a counter example? I can't think of any!
     
  2. jcsd
  3. Oct 5, 2012 #2

    Mark44

    Staff: Mentor

    If the sequence {an2} converges, then for any ε > 0, there is a number n1 such that |an2 - L2 | < ε when n >= n1. Given that the sequence {an} converges, can you use this to show that {an2} also converges?
     
  4. Oct 5, 2012 #3
    You can also start from:
    [tex]|a^2_n-L^2|=|a_n+L|\cdot|a_n-L|[/tex]
     
  5. Oct 5, 2012 #4

    SammyS

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    I'm quite sure Mark means, |an2 - L2 | < ε ...
     
  6. Oct 5, 2012 #5

    Mark44

    Staff: Mentor

    No, I meant L2, to distinguish it from L. It might turn out that L2 = L2, but I didn't want to make that assumption.
     
  7. Oct 7, 2012 #6
    [tex]
    \forall \epsilon>0\ \exists N_1: n >N_1\ |a_n -L|< \epsilon[/tex]

    but also

    [tex]
    \forall \epsilon>0\ \exists N_2: n >N_2\ |a_n -L|< \epsilon_1 = \frac{\epsilon}{M+L}[/tex]

    Remember that every convergent sequence is bounded from above, say in this case by M.

    Now we get:

    [tex] |a_n^2 -L^2| = |a_n-L||a_n+L| <\epsilon_1 (M+L)<\epsilon [/tex] if [tex]n>max(N_1,N_2) [/tex]
     
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