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Math Help - Analysis: Please check my proof on the arithmetic of CONVERGENT SEQUENCES

  1. #1
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    Analysis: Please check my proof on the arithmetic of CONVERGENT SEQUENCES

    I am new to analysis and am having a lot of trouble with this proof. I am not sure if what I have written is valid. Please let me know. Thank you.

    Theorem: If {an} converges to A and {an + bn} converges to A + B, then {bn} converges.


    Proof:

    Since {an + bn} converges to A + B, for every ε> 0 there exists an N such that for
    n > N, |(an + bn) - (A + B)| < ε

    =>
    |(an - A) + (bn - B)| < ε
    It is true that
    |(an - A) + (bn - B)| ≤ |an - A| + |bn - B|
    It is also true that
    |(an - A) + (bn - B)| - |an - A| < ε for n > N
    Thus it is true that
    |(an - A) + (bn - B)| - |an - A| ≤ |an - A| + |bn - B| - |an - A| < ε
    for n > N
    Therefore
    |an - A| + |bn - B| - |an - A| < ε for n > N
    so
    |bn - B| < ε for n > N
    Since for ε there exists an N such that n > N => |bn - B| < ε, then {bn} converges.■





    I used the same N for {bn} that I used for {an +bn} because if it works for
    {an + bn} and {bn} is inside {an + bn}, then I reason it will work for {bn} too.
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  2. #2
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    First, one can never assume that same N works in all cases.

    Now you know that |W|\le |W-U|+|U|.
    Let U=A-a_n~\&~W=b_n-B.
    You can controle both |A-a_n|~\&~|a_n+b_n-(A+B)| making each less than \frac{\epsilon}{2}.
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  3. #3
    Member HappyJoe's Avatar
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    Do you already know the theorems that if \{x_n\} and \{y_n\} converges, then \{cx_n\} converges for any constant c, and \{x_n+y_n\} converges?

    In that case, you can take a short-cut. Since \{a_n\} converges, you know that \{-a_n\} converges. Then, since \{a_n+b_n\} converges, you know that \{a_n+b_n+(-a_n)\} = \{b_n\} converges.
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  4. #4
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    Quote Originally Posted by HappyJoe View Post
    Do you already know the theorems that if \{x_n\} and \{y_n\} converges, then \{cx_n\} converges for any constant c, and \{x_n+y_n\} converges?

    In that case, you can take a short-cut. Since \{a_n\} converges, you know that \{-a_n\} converges. Then, since \{a_n+b_n\} converges, you know that \{a_n+b_n+(-a_n)\} = \{b_n\} converges.

    Is that essentially letting \{a_n+b_n\} = \{x_n\}, and \{-a_n\} = \{y_n\} in the sequence \{x_n+y_n\}, which is known by theorem to converge?


    Quote Originally Posted by Plato View Post
    First, one can never assume that same N works in all cases.

    Now you know that |W|\le |W-U|+|U|.
    Let U=A-a_n~\&~W=b_n-B.
    You can controle both |A-a_n|~\&~|a_n+b_n-(A+B)| making each less than \frac{\epsilon}{2}.

    My question about this is that I already know that in the proof where you're given

    \{a_n\} converges to A and \{b_n\} converges to B, prove that \{a_n+b_n\} converges to A + B

    you use the exact same thing:

    |a_n- A| < ε/2

    and

    |b_n-B| < ε/2


    So if in this proof |b_n-B| < ε/2, how can |a_n+b_n-(A+B)| also be less than ε/2 for the proof I'm trying to understand in the original post? Because isn't that essentially saying that all three sequences are less than ε/2?

    Well, what I mean is, I see how all three could, since epsilon is arbitrary, but I don't see how it helps me. (I do realize this is because of me, not you, so don't get the wrong idea. I'm just a little lost here.)





    Thanks for both of your respective input!
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  5. #5
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    If (a_n)\to A~\&~(a_n+b_n)\to A+B then we have.
    \left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n  - A} \right| < \varepsilon /2} \right] and \left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {\left( {a_n  + b_n } \right) - \left( {A + B} \right)} \right| < \varepsilon /2} \right]

    Now let N=N_1+N_2 if n\ge N then n>N_1~\&~n>N_2 so both of the above conditions are true.

    Can you finish?
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  6. #6
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    Quote Originally Posted by Plato View Post
    If (a_n)\to A~\&~(a_n+b_n)\to A+B then we have.
    \left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n  - A} \right| < \varepsilon /2} \right] and \left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {\left( {a_n  + b_n } \right) - \left( {A + B} \right)} \right| < \varepsilon /2} \right]

    Now let N=N_1+N_2 if n\ge N then n>N_1~\&~n>N_2 so both of the above conditions are true.

    Can you finish?
    I don't know. I'll try. If you don't mind, criticism would be nice. I know that I suck, so hopefully I won't infuriate you with my denseness.


    Since N=N_1+N_2, then

    |a_n- A| + |a_n+b_n-(A+B)| <\varepsilon /2  +  \varepsilon /2  = \varepsilon

    So then

    |a_n- A| + |a_n+b_n-(A+B)| |a_n- A| + |a_n- A| + |b_n- B| <\varepsilon /2  +  \varepsilon /2  = \varepsilon

    So

    |a_n- A| + |a_n+b_n-(A+B)| 2|a_n- A| + |b_n- B| <\varepsilon /2  +  \varepsilon /2  = \varepsilon

    This should be satisfied when

    |b_n- B| <\varepsilon/2

    and


    |a_n- B| <\varepsilon/2

    but since we already know
    |a_n- B| <\varepsilon/2

    then

    |b_n- B| <\varepsilon/2

    for

    n\ge N


    therefore

    \{b_n\} converges.
    Last edited by SunRiseAir; October 25th 2010 at 04:38 PM.
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  7. #7
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    Quote Originally Posted by SunRiseAir View Post
    Is that essentially letting \{a_n+b_n\} = \{x_n\}, and \{-a_n\} = \{y_n\} in the sequence \{x_n+y_n\}, which is known by theorem to converge?
    Yeah, that's exactly it.
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