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Proof on Sequences: Sum of a convergent and divergent diverges

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove if sequence [itex]a_{n}[/itex] converges and sequence [itex]b_{n}[/itex] diverges, then the sequence [itex]a_{n}[/itex]+[itex]b_{n}[/itex] also diverges.

    2. Relevant equations



    3. The attempt at a solution

    My professor recommended a proof by contradiction. That is, suppose [itex]a_{n}[/itex]+[itex]b_{n}[/itex] does converge. Then, for every ε > 0, there exists a natural number [itex]N_{1}[/itex] so that n > [itex]N_{1}[/itex] implies |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|< ε

    We already know there exists [itex]N_{2}[/itex] so that n > [itex]N_{2}[/itex] implies |[itex]a_{n}[/itex] - M| < ε. So let N = max{[itex]N_{1}[/itex], [itex]N_{2}[/itex]}. Then n > N means we know [itex]a_{n}[/itex] is "very close" to M. My purpose in this is to try and show that this implies [itex]b_{n}[/itex] has a limit (that is, it converges) providing a contradiction. However, I'm not sure how to go about this.
     
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  3. Oct 10, 2011 #2

    SammyS

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    That's the general idea !

    Rewrite [itex]\left|a_n+b_n-L\right|[/itex] as [itex]\left|a_n-M+b_n-(L-M)\right|\,,[/itex] then use the triangle inequality.
     
  4. Oct 10, 2011 #3
    Ah! I think I've got it.

    Proof: Assume [itex]a_{n}[/itex] is convergent and [itex]b_{n}[/itex] is divergent.

    Now suppose that [itex]a_{n}[/itex]+[itex]b_{n}[/itex] is convergent.
    Then [for every ε > 0 there exists a natural number [itex]N_{1}[/itex] so that
    n > [itex]N_{1}[/itex] implies |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|< ε/2

    We know by our assumption that there also exists natural number [itex]N_{2}[/itex] so that
    n > [itex]N_{2}[/itex] implies |[itex]a_{n}[/itex]-M| < ε/2

    Now let N = max{[itex]N_{1}[/itex],[itex]N_{2}[/itex]}
    Then n > N implies
    |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|< ε/2 and
    |[itex]a_{n}[/itex]-M| < ε/2
    So |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|+ |-([itex]a_{n}[/itex]-M)| < ε
    and |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - [itex]a_{n}[/itex] - (L-M)| < ε
    (by the triangle inequality)
    But [itex]a_{n}[/itex]+[itex]b_{n}[/itex] - [itex]a_{n}[/itex] = [itex]b_{n}[/itex]
    so this last statement implies |[itex]b_{n}[/itex] - (L-M)| < ε
    which implies [itex]b_{n}[/itex] converges, which is a contradiction
    Therefore, [itex]a_{n}[/itex]+[itex]b_{n}[/itex] diverges
     
  5. Oct 10, 2011 #4

    SammyS

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    Make it clear that by the triangle ineq. [itex]\left|a_n+b_n-L-a_n+M\right|\le\left|a_n+b_n-L\right|+\left|-a_n+M\right|<\varepsilon[/itex]
     
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