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Converging analysis proof

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume that (an) is a bounded (but not necessarily convergent) sequence, and that the
    sequence (bn) converges to 0. Prove that the sequence (anbn) converges to zero.

    2. Relevant equations



    3. The attempt at a solution

    Assume that an is a bounded sequence and bn converges to 0.

    That means for all n in N, there exists a M >0 so that
    |an|<=M
    Since bn converges, that means that it must be bounded as well. Which means for all n in N there exists a P>0 so that
    |bn|<=P

    since |an|<=M and |bn|<=P that means for all n in N:
    |an||bn|<= MP which is equivalent to |anbn|<=MP
    where MP>0 since M>0 and P>0. Hence (anbn) is bounded

    Since bn converges to 0 that means for e>0 there exists an N in N so that for n>=N
    |bn-0|<e
    which is equivalent to -e<bn<e

    This is where I get stuck. Do I just multiply the inequality by an? cause then I'd have
    -e(an)<bnan<e(an)
    which would be equivalent to |anbn|<e2 if I let e2=e(an) which would mean that anbn converges to zero as well. But I don't know if I can multiply the sequence by it though....

    Any help would be great!
     
  2. jcsd
  3. Sep 26, 2009 #2

    lanedance

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    Homework Helper

    an bounded look alright
    As bn is convergent, I would say for any P>0, there exists N such that for all n>N then
    |bn-0|< |bn|


    i think you were almost there...

    now what you need to show to prove an.bn converges to zero, is that for any e>0 you can choose N, such that for all n>N you have
    |an.bn|<e

    as you know an<=M for all n, then
    |an.bn|<=|M.bn|

    so now you just need to show you can choose N such that for all n>N
    |bn|<=e/|M|
    and i think you're there
     
  4. Sep 26, 2009 #3

    lanedance

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    updated above
     
  5. Sep 27, 2009 #4
    thanks for the help :)
     
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