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Math Help - {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

  1. #1
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    {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    \{a_n\} and \{a_n+b_n\} converge. Prove \{b_n\} converges.

    Let \{a_n\} and \{a_n+b_n\} converge to A, \ A+\alpha, respectively,

    Then there is an \epsilon>0 and \exists N\in\mathbb{N}, \ n\geq N,

    |a_n-A|<\frac{\epsilon}{2}

    and

    |a_n+b_n-(A+\alpha)|\leq \epsilon

    |a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon

    \Rightarrow |b_n-\alpha|<\frac{\epsilon}{2}<\epsilon

    Therefore, \{b_n\} converges.

    Correct?
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  2. #2
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    It looks good except for this part:

    Then there is an \epsilon>0
    You need to say something like "Let \epsilon>0 be given."
    Otherwise you are saying that there exists an epsilon and N for which the following is true, which is not what you want to prove. By saying that epsilon is given and arbitrary, you essentially show that it is true for any epsilon (greater than zero, of course).
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    As Joanna said, it needs to be a "for all epsilon", not a "there exists an epsilon". You need to work with

    |b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.

    See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    Quote Originally Posted by Ackbeet View Post
    As Joanna said, it needs to be a "for all epsilon", not a "there exists an epsilon". You need to work with

    |b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.

    See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.
    What is "not valid"?

    We know \{a_n+b_n\} converges. So if we let it converge to A+\alpha, we have |a_n+b_n-(A+\alpha)|<\epsilon

    And by the triangle inequality, |a_n-A|+|b_n-\alpha|\geq |a_n+b_n-(A+\alpha)|

    Since we also know \{a_n\} converges, I can let it converge to some A.

    Because it converges |a_n-A|<\frac{\epsilon}{2}<\epsilon

    Since |a_n-A|<\frac{\epsilon}{2}, we can substitute \frac{\epsilon}{2} for |a_n-A| and make the inequality strictly less than.

    That yields:

    |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|

    But we also know that |a_n+n_n-(A+\alpha)|<\epsilon.

    So \frac{\epsilon}{2}+|b_n-\alpha|<\epsilon

    I don't see what is wrong in the inequality structure.
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    Quote Originally Posted by Ackbeet View Post
    |b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.

    See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.
    Right! Missed that for some reason.


    Quote Originally Posted by dwsmith View Post
    \{a_n\} and \{a_n+b_n\} converge. Prove \{b_n\} converges.

    |a_n+b_n-(A+\alpha)|\leq \epsilon

    |a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon
    You cannot say that \frac{\epsilon}{2}+|b_n-\alpha|<\epsilon. You know that |a_n+b_n-(A+\alpha)|<\epsilon, but you just showed that |a_n+b_n-(A+\alpha)|<\frac{\epsilon}{2}+|b_n-\alpha| so it is possible that \frac{\epsilon}{2}+|b_n-\alpha|\geq\epsilon.
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    I don't agree with you two and here is why:

    So if we let \epsilon>0. Then \exists N\in\mathbb{N}, \ n\geq N

    Now, let \{a_n\}\to A \ \text{and} \ \{a_n+b_n\}\to A+\alpha

    Since those two sequences converge,

    |a_n-A|<\epsilon \ \text{but it is also true then that} \ |a_n-A|<\frac{\epsilon}{2}<\epsilon

    And

    |a_n+b_n-(A+\alpha)|<\epsilon

    By the triangle inequality,

    |a_n+b_n-(A+\alpha)|\leq \underbrace{|a_n-A|}_{|a_n-A|<\frac{\epsilon}{2}}+|b_n-\alpha|<\underbrace{\frac{\epsilon}{2}}_{\text{thi  s is + and} \ < |a_n-A|}+|b_n-\alpha|<\epsilon

    Now, we have

    |a_n+b_n-(A+\alpha)|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon

    Let's subtract epsilon over 2.

    |b_n-\alpha|<\frac{\epsilon}{2}

    I started with something giving to be true {a_n+b_n} converges, used another true piece {a_n} converges, used the triangle inequality, and made substitution that is true.

    If |a_n+b_n-(A-\alpha)|\leq|a_n-A|+|b_n-\alpha|, it is <\frac{\epsilon}{2}+|b_n-\alpha| which is by definition less than epsilon.
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    \underbrace{|a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|}_{\text{I agree with all of this,}}<\epsilon.

    |a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\underbrace{\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon}_{\text{but not this.}}.

    You're essentially saying that because A < B and A < C, that therefore A < B < C. But 4 < 5, and 4 < 6; that does not imply 4 < 6 < 5.
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    Quote Originally Posted by dwsmith View Post
    \{a_n\} and \{a_n+b_n\} converge. Prove \{b_n\} converges.
    I know this has been answered. But here is a flowing proof.
    First suppose that (a_n)\to\alpha~\&~(a_n+b_n)\to \alpha+\beta.
    Suppose that \varepsilon  > 0. Now use the definition twice.
    \left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n  - \alpha } \right|} \right] < \frac{\varepsilon }{2}
    \left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {a_n  + \beta _n  - (\alpha  + \beta) } \right|} \right] < \frac{\varepsilon } {2}.

    Note that we need two values of N that we need to insure both condition hold.
    So let N=N_1+N_2. If n\ge N then
    \begin{array}{rcl}{\left| {b_n  - \beta } \right|} &  =  & {\left| {\left( {a_n  + \beta _n } \right) - \left( {\alpha  + \beta } \right) + \alpha  -a_n } \right|}\\{} &  \leqslant  & {\left| {\left( {a_n  + \beta _n } \right) - \left( {\alpha  + \beta } \right)} \right| + \left| {\alpha  - a_n } \right|}\\{} &  \leqslant  & {\frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon }  \\ \end{array}
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    Re: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

    Here's a specific example where your argument fails.

    As Plato pointed out, you need an N for each sequence; let N_{1} be such that n>N_{1} implies |a_{n}-A|<\frac{\epsilon}{2} and let N_{2} be such that n>N_{2} implies |a_{n}+b_{n}-(A+\alpha)|<\epsilon. Then we can obtain both inequalities with N=\max\{N_{1}, N_{2}\}.
    Then n>N implies |a_{n}-A|<\frac{\epsilon}{2} and |a_{n}+b_{n}-(A+\alpha)|<\epsilon. This is a particular N.

    Suppose that we already know that b_{n} converges to \alpha. Also suppose that \alpha - b_{N+1}=\epsilon and a_{N+1}-A=\frac{\epsilon}{3} (which is certainly less than \frac{\epsilon}{2}).

    Then we have
    |a_{N+1}+b_{N+1}-(A+\alpha)|=|a_{N+1}-A-(\alpha - b_{N+1})|=|\frac{\epsilon}{3}-\epsilon|=|-\frac{2 \epsilon}{3}|<\epsilon but
    |a_{N+1}-A|+|b_{N+1}-\alpha|=\frac{\epsilon}{3}+\epsilon=\frac{4 \epsilon }{3}>\epsilon.

    The point is that there exists an n_{1}>N such that |b_{n_{1}}-\alpha|=\epsilon and |a_{n_{1}}+b_{n_{1}}-(A+\alpha)|<\epsilon. You would need to find a larger N to get |b_{n}-\alpha|<\epsilon.
    Last edited by Joanna; June 24th 2011 at 02:10 PM. Reason: Was not displaying an epsilon..
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