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Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} converges

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose {an}n=1 and {bn}n=1 are sequences such that {an}n=1 and {an + bn }n=1 converge.

    Prove that {bn}n=1 converges.



    2. Relevant equations

    The definition of convergence.


    3. The attempt at a solution

    I am pretty new to mathematics that requires proof, so excuse me if I do something really stupid... but basically, is this a sufficient proof?


    1. Assume {an}n=1 converges to A (by hypothesis).

    Then for ε/2 > 0 there is a positive integer N1 such that if n ≥ N1, then |an - A| < ε/2.​

    2. Assume that {an + bn }n=1 converges to A + B (by hypothesis).

    Then for ε > 0 there is a positive integer N = max{N1, N2} such that if n ≥ N, then | (an + bn) - (A + B) | < ε​

    3. | (an + bn) - (A + B) | = | (an - A) + (bn - B) | < ε

    4. Since by hypothesis |an - A| < ε/2, then

    | (an - A) - (an - A) + (bn - B) | < ε - ε/2

    | (bn - B) | < ε/2

    if n ≥ N2 for some positive integer N2.​

    5. But this is the definition of convergence, therefore {bn}n=1 converges (to B).



    Thanks.
     
    Last edited: Sep 22, 2010
  2. jcsd
  3. Sep 22, 2010 #2

    Office_Shredder

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    Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

    You never actually say what N2 is.

    Also, if a+b<e and a<e/2 that doesn't give you that b<e/2, so I'm not sure where that final inequality comes from. Subtracting [tex]|a_n-A|[/tex] from the left hand side does not allow you to just move it inside the absolute value sign; and subtracting |a_n-A| from the right hand side, you can't replace it with [tex]\epsilon/2[/tex] and maintain the same inequality, since that makes the right hand side smaller, not larger



    As a fast way to see a lot of results like this, once you have the standard summation and multiplication rules, you can use

    [tex]b_n=(a_n+b_n)-a_n[/tex] and use what you know about the summation of sequences
     
    Last edited: Sep 22, 2010
  4. Sep 22, 2010 #3

    vela

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    Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

    There's a problem in step 4 of the attempt. It doesn't follow from

    | (an - A) + (bn - B) | < ε

    that

    | (an - A) - (an - A) + (bn - B) | < ε - ε/2

    It's like saying |1-1.9|=0.9 < 1 so |1-1-1.9| < 1-1=0.
     
  5. Sep 22, 2010 #4

    hunt_mat

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    Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

    This looks okay, why not look at the algebra of limits? if [tex]a_{n}+b_{n}\rightarrow b}[/tex] and [tex]a_{n}\rightarrow a[/tex] then the sequence [tex]b_{n}=a_{n}+b_{n}-a_{n}\rightarrow b-a[/tex]
     
  6. Sep 23, 2010 #5
    Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

    Thanks everyone. I'm clearly missing something in the understanding of this material, so I'll take what you've said this weekend and dig through the book and see if I can spot the misunderstanding.
     
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