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Real Analysis: Proving an*bn converges to ab

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the fact that an= a + (an - a) and bn= b + (bn - b) to establish the equality an*bn - ab = (an-a)(bn-b)+b(an-a)+a(bn-b).

    Then, use this equality to prove that the sequence {an*bn} converges to ab.


    2. Relevant equations
    Definition of convergence: |an*bn - ab| < ε


    3. The attempt at a solution

    The first part was easy; just basic algebra. I'm stuck on the last part. I'm not sure where to begin. I tried expanding out the right side, hoping to find something I could use the triangle inequality on. I ended up with

    an*bn - ab= an*bn -b*an -a*bn +ab+b*an-ab+a*bn-ab

    = an*bn+a*bn-ab

    = bn(an+a)-ab

    I don't think I can do anything with that. Any suggestions where I can go from here? Am I even starting in the right place?
     
  2. jcsd
  3. Apr 8, 2012 #2
    Is the problem statement: if [itex]\displaystyle \lim_{n\to \infty} a_n = a[/itex] and [itex]\displaystyle \lim_{n\to \infty} b_n = b[/itex] proove that [itex]\displaystyle \lim_{n\to \infty} a_nb_n = ab[/itex]??
     
  4. Apr 8, 2012 #3
    Yes, sorry. Assume that an and bn converge to a and b, respectively.
     
  5. Apr 8, 2012 #4

    Office_Shredder

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    All you need to do is show that (an-a)(bn-b)+b(an-a)+a(bn-b) goes to zero as n goes to infinity. If n is really big, what can you say about an-a? What about bn-b? What does that say about the size of the entire expression?

    The triangle inequality you might want to use is
    [tex] |(a_n-a)(b_n-b)+b(a_n-a)+a(b_n-b)| \leq |(a_n-a)(b_n-b)|+|b(a_n-a)|+|a(b_n-b)|[/tex]
     
  6. Apr 8, 2012 #5
    Ya I agree with the above posters. This type of question is a triangle inequality question.
     
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