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Short proof of sequences

  1. Nov 4, 2013 #1
    Prove that if (a_n) is a null sequence and (b_n) is a bounded sequence then the sequence (a_nb_n) is null:

    from definitions if b_n is bounded then ## \exists H \in \mathbb{R} ## s.t. ## |b_n| \leq H ## if a_n is a null sequence it converges to 0 (from my book), i.e. given ## \epsilon ' > 0 ## ## \exists N \in \mathbb{R} ## s.t. n>N ## \Rightarrow |a_n| < \epsilon ' ## set ## e' = \dfrac{\epsilon}{|H|} ## then for ## n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon ## i.e. a_nb_n also converges to 0.

    I'm slightly worried about this, if H is negative, then can we just multiply ## |a_n| ## and ## |b_n| ## as afaik from the axioms if ## a>0 ## ## x>y## then ## xa>ya ## and I'm not sure if H is negative or if |a_n| is negative if we can simply just multiply the two inequalities. Any help, explaining this, thanks.
     
  2. jcsd
  3. Nov 4, 2013 #2

    LCKurtz

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    How can ##|a_n|## be negative?
     
  4. Nov 4, 2013 #3

    LCKurtz

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    Since your proof is correct, I think I can get away with showing you how to write it up a bit nicer:

    Since ##\{b_n\}## is bounded there is ##H \in \mathbb{R} ## such that ##|b_n|\le H## for all natural numbers ##n##. Suppose ##\epsilon > 0##. Since ##a_n\to 0## there exists ##N\in \mathbb R## such that if ##n>N## we have ##|a_n| < \frac \epsilon H##. Now if ##n>N## then we have ##|a_nb_n-0| = |a_n||b_n| < \frac \epsilon H\cdot H =\epsilon##. Hence ##a_nb_n\to 0##.
     
  5. Nov 5, 2013 #4
    ok ##|a_n| < \epsilon ## this is the distance from a_n to 0 which is less than epsilon so I guess it can't be negative. But what if ## |b_n| \leq H ## and H is negative? take e.g. ## |b_n| \leq -1 ## then ## 1 \leq b_n \leq -1 ##
     
  6. Nov 5, 2013 #5

    LCKurtz

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    Doesn't ##1\le -1## bother you? Or ##|b_n|\le -1##?
     
  7. Nov 5, 2013 #6
    ha - yeah, thanks.

    Also thanks for writing up a more concise proof!
     
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