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Homework Help: Lim a(n)b(n) = AB

  1. Feb 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Theorem. If lim a n = A and lim b n = B, then lim a n b n = AB.

    2. Relevant equations

    |an - A| < ε

    |bn - B| > ε

    3. The attempt at a solution

    I have the solution to this, but I'm unclear on why one part is done.

    Let ε > 0. Since bn converges it is bounded and there exists an M1 such that for all n [itex]\in[/itex] N we have |bn| < M1. Define M1 [itex]\geq[/itex] 1. Then |bn| / M1 ≤ 1.

    Skipping ahead, do the same for an and let |an - A| < [itex]\frac{ε}{2M1}[/itex]

    and let

    |bn - B| < [itex]\frac{ε}{2M2}[/itex].

    |anbn - AB| = |anbn -Abn + Abn - AB|

    So are we trying to manipulate the limit anbn into the familiar definition of |an - L| < ε so we can compare them?

    So |anbn -Abn + Abn - AB|

    |anbn -Abn| + |Abn - AB|


    |an - A||bn| + |bn - B||A|





    Stupid question, but how are we multiplying each ε by |bn| and |A|? I know that it gives us the two original lim an and lim bn, but how can we just stick them on the other side of the "<" symbol? And how do |bn|/M1 = 1 and |A|/M2 = 1 if M is just defined as being larger than both? Can we just take the max of |bn| and |A|?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 19, 2014 #2


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    (In LaTeX, don't use the special characters at the right, like "ε" . Use \varepsilon . For subscripts use M_1, etc.)

    First it will help to state what is required to show that ##\displaystyle \lim\ \left(a_n\ b_n\right) = AB \ ## .

    You know that the sequences an and Bn converge, so you can use any positive quantity to represent "ε" for each of these sequences.
  4. Feb 20, 2014 #3
    |an - A||bn| + |bn - B||A|

    You know that bn converges thus there is a M such that bn<M for all n. Since |an-A| can be made really small there is a n such that for all n> N there holds

    |an - A| < epsilon/(2*(M+1)).

    but then for all n>N:

    |an - A||bn| < epsilon*M/(2*(M+1)) < epsilon /2.
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