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Homework Help: {Sn} is convergent -> {|Sn|} is convergent

  1. Nov 6, 2011 #1
    {Sn} is convergent ---> {|Sn|} is convergent

    1. The problem statement, all variables and given/known data

    I need to prove that if {sn} is convergent, then {|sn|} is convergent.

    2. Relevant equations

    sn is convergent if for some s and all ε > 0 there exists a positive integer N such that |sn - s| < ε whenever nN.

    3. The attempt at a solution

    Proof. By contrapositive. Suppose {|sn|} is not convergent. Then for all s there exists an ε > 0 such that ||sn| - s|| ≥ ε for all n.

    ..... I need to somehow show that this implies that {sn} does not converge.

    Maybe some fancy triangle inequality thing like

    ε ≤ ||sn| - s| ≤ ||sn| - sn| + |sn - s|

    Wat do, PF?
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2
    Re: {Sn} is convergent ---> {|Sn|} is convergent

    Ah, wait. I think I figured out something.

    ε ≤ ||sn| - s| ≤ ||sn| - sn| + |sn - s| ≤ 2|sn| + |sn - s|

    Am I any closer?
     
    Last edited: Nov 6, 2011
  4. Nov 6, 2011 #3
    Re: {Sn} is convergent ---> {|Sn|} is convergent

    You are supposed to show that |Sn| converges to |S| not to S.
    The trick is to show that ||Sn|-|S|| <= |Sn-S|

    Hint to showing it:
    |S| = |Sn + (S-Sn)|
    And use the triangle inequality.
     
  5. Nov 6, 2011 #4
    Re: {Sn} is convergent ---> {|Sn|} is convergent

    I knew the triangle inequality would show his face somewhere.
     
  6. Nov 6, 2011 #5
    Re: {Sn} is convergent ---> {|Sn|} is convergent

    screen-capture-57.png



    http://images.icanhascheezburger.com/completestore/2009/3/5/128807888821558038.jpg
     
  7. Nov 6, 2011 #6
    Re: {Sn} is convergent ---> {|Sn|} is convergent

    Hey, I need a hint for another problem. (You guys give stellar hints. Pat yourselves on the back.)

    Suppose {pn} is a Cauchy sequence in a metric space X, and some subsequence {pni} converges to a point p in X. Prove that the full sequence {pn} converges to p.

    As always, I'd love to use contradiction.

    Suppose {pn} does not converge to p. Then for some ε > 0, d(p,pn) ≥ ε for all n. Then d(pm,pn)+d(pm,p) ≥ ε for all m, n. But {pm} is a Cauchy sequence, so with the proper choice of m, n we'll have d(pm,pn) < ε. ........ blah blah I hope this is going somewhere. A little hint, maybe?
     
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