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Math Help - sum of 2 convergent sequences

  1. #1
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    sum of 2 convergent sequences

    Defn: Pn converges to point p means that if S is an open interval containing p then there is a positive integer N such that if n is a positive integer and n>N then Pn is in S.

    I need to show that if the Pn converges to p and Qn converges to q then Pn+Qn converges to p+q.

    I feel like I'm close to getting it, but not quite there yet, so any help would be greatly appreciated.

    Note: I can't use the epsilon proof; I must use intervals.
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  2. #2
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    Hmm....I understand that almost all the terms of Pn and Qn will be between a+c and b+d, but how can I show that they converge to p+q?
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  3. #3
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    That is just another way of saying ‘converges’. That is, if each open interval that contains p also contains ‘almost all’ [all but a finite collection] of the terms of \left( {p_n  } \right) then the sequence converges to p.

    I deleted the post because there was an error. The interval was not general enough.
    So here is a correction. Suppose that \left( {\exists (a,b)} \right)\left[ {p + q \in (a,b)} \right].
    Last edited by Plato; September 24th 2009 at 02:51 PM.
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  4. #4
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    Still not clear...
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  5. #5
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    Quote Originally Posted by spikedpunch View Post
    Still not clear...
    Well think about how you do this with epsilon-neighborhoods.
    Then adapt that to open intervals.
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  6. #6
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    This is what I have so far.

    p is in (a,b) and q is in (c,d). let epsilon1= min (p-a,p+b) and epsilon2=min(q-c,q+d).
    For some N1, n>N1, -epsilon1<Pn-p<epsilon1 and
    for some N2, n>N2, -epsilon2<Qn-q<epsilon2.
    Let N=max(N1,N2).

    Now, I want to show that Pn+Qn converges to (p+q), but I'm not sure if simply combining the inequalities will show this.

    Apologies for repeated questions on the same problem, but I'm having a difficult time wrapping my head around this convergence stuff.
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  7. #7
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    You must start with p+q\in (a,b).
    So \left( {\exists \varepsilon  > 0} \right)\left[ {\left( {p + q - \varepsilon ,p + q + \varepsilon } \right) \subset (a,b)} \right]
    Now consider two open intervals: O_p  = \left( {p - \frac{\varepsilon }{2},p + \frac{\varepsilon }{2}} \right)\;\& \;O_q  = \left( {q - \frac{\varepsilon }{2},q + \frac{\varepsilon }{2}} \right)
    It should be clear that p\in O_p\;\& \;q\in O_q.
    If x\in O_p\;\& \;y\in O_q can you show that x+y\in (a,b)?
    Last edited by Plato; September 25th 2009 at 10:10 AM.
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  8. #8
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    I don't see how the sequences Pn and Qn fit in to what you are saying.
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  9. #9
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    Quote Originally Posted by spikedpunch View Post
    I don't see how the sequences Pn and Qn fit in to what you are saying.
    Almost all of the terms of \left(p_n\right) are in O_p by convergence.
    Almost all of the terms of \left(q_n\right) are in O_q by convergence.
    Therefore Almost all of the terms of \left(p_n+q_n\right) are in (a,b)
    Because p+q\in(a,b) that is sufficient to prove that \left(p_n+q_n\right)\to(p+q)
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  10. #10
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    I can't use the phrase "almost all the terms" in my proof. I can only use the definition given in my initial post.

    Again, apologies.
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