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Homework Help: Let's suppose that both sequences are convergent

  1. Dec 22, 2004 #1
    Consider the following statement:

    If [tex] \left\{ a_n \right\} [/tex] and [tex] \left\{ b_n \right\} [/tex] are divergent, then [tex] \left\{ a_n b_n \right\} [/tex] is divergent.

    I need to decide whether it is true or false, and explain why. The real problem is that I checked the answer in my book; it's false, but I don't understand it. Here is what I think:

    Let's suppose that both sequences are convergent. Then, it follows that

    [tex] \lim _{n\to \infty} a_n \cdot \lim _{n\to \infty} a_n = \lim _{n\to \infty} \left( a_n b_n \right) \tag{1} [/tex]​

    But, the truth is that both are divergent. So, [tex] \lim _{n\to \infty} a_n \neq 0 [/tex] and [tex] \lim _{n\to \infty} b_n \neq 0 [/tex]. If neither is zero, then how can [tex] \lim _{n\to \infty} \left( a_n b_n \right) = 0[/tex] (so that the statement is false)? It doesn't sound reasonable if you consider (1).

    Can anybody please help me clarify this?

    Thank you very much.
  2. jcsd
  3. Dec 22, 2004 #2


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    If a sequence is convergent, it doesn't decessarily means it converges towards 0, like you seemed to be implying in your post.

    To decide wheter the proposition it's true or false, a simple counter exemple suffice. Consider [itex]a_n = (-1)^n[/itex] and [itex]b_n = (-1)^n[/itex]. These are both divergent series because when n is pair, [itex]a_n = 1[/itex] and when n is odd [itex]a_n = -1[/itex], such that the limit is dependent upon n ==> it is not unique ==> it doesn't exist ==> the sequences diverge. But [itex]a_n b_n = (-1)^{2n} = 1 \ \forall n \in \mathbb{N}[/itex] is a sequence that converges towards 1.

    Or take [itex]a_n = (-1)^n[/itex] and [itex]b_n = (-1)^n+1[/itex]. Then [itex]a_n b_n = (-1)^{2n+1} = -1 \ \forall n \in \mathbb{N}[/itex], which converges towards -1.
    Last edited: Dec 22, 2004
  4. Dec 22, 2004 #3


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    N.B. But in the case where [itex]a_n[/itex] and [itex]b_n[/itex] are divergent because they increase or decrease without limit (i.e. because their limit is plus or minus infinity), then it is true that [itex]a_n b_n[/itex] is also a divergent sequence.
  5. Dec 22, 2004 #4
    Oh... I see what you mean. I thought that way because I had in mind the theorem that says that when a series is convergent, the terms go to 0. It doesn't apply in this case, since we only have sequences. Thanks.
  6. Dec 22, 2004 #5
    I see, so it isn't true in general.
  7. Dec 22, 2004 #6


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    The following is a problem step:
    The hypothesis is that both sequences are divergent, not convergent, so this thought does not apply to this question in any useful way.

    For example, [itex]a_n=b_n=(-1)^n[/itex] are two divergent sequences, but [itex]\{a_nb_n\}[/itex] is constant, so it clearly converges.

    By the way, the theorem about series on indicates that terms in a convergent series go to zero. There are divergent series that go to zero like the harmonic series:
    [tex]\sum_{i=1}^{\infty} \frac{1}{i}[/tex]
    And, since
    [tex]\sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6}[/tex]
    is convergent, the above is false for series as well as sequences.
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