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Homework Help: Apostol Calculus Vol1 10.4.29 Sequences

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Assume that [tex] \{ a_n\}\rightarrow 0 [/tex]. Use the definition of limit to prove that [tex] \{ a_n^2\} \rightarrow 0[/tex].

    2. Relevant equations

    Definition of limit. For all ε>0 there exists N s.t. n>N implies |a_n - L|<ε.

    3. The attempt at a solution

    I know why this is true... if the sequence goes to zero then a_n<1. Therefore [tex] a_n^2 < a_n < 1 [/tex]. Then [tex] a_n^2 [/tex] is bounded above by a_n and below by zero, so it also converges. Is this as simple as:

    Assume that given an ε>0 we choose N s.t. for all n>N implies |a_n - 0|<sqrt(ε). Then since [tex] a_n > a_n^2[/tex] we have [tex] |a_n^2|<\varepsilon [/tex]. Where L=0.
    Last edited: Oct 21, 2012
  2. jcsd
  3. Oct 21, 2012 #2


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    Not quite. You weren't given ##a_n## nonnegative, nor is it true for all ##n##. There exists ##N>0## such that ##|a_n|<1## if ##n>N##.

    ##a_n^2 < |a_n|<1## if ##n>N##.

    Yes, it is pretty much that simple. Just write it up carefully.
  4. Oct 21, 2012 #3
    Okay, so how about:

    Suppose that [tex] \{ a_n\}\rightarrow 0.[/tex] Then given ε>0 there exists N s.t. [tex] |a_n|<\sqrt{\varepsilon } [/tex] whenever n>N. By hypothesis [tex] |a_n| < 1 .[/tex] Therefore [tex] a_n^2 < |a_n| < 1[/tex] and [tex] |a_n|<\sqrt{\varepsilon } \Longrightarrow |a_n^2| < \varepsilon .[/tex] Therefore [tex] \{ a_n^2\} \rightarrow 0. [/tex]
  5. Oct 21, 2012 #4


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    Given the way you have started, which is good, why don't you just leave out everything in red?
  6. Oct 21, 2012 #5
    Ah I see. I am used to being overly detailed in explanations because my proof based class is a lower-div linear algebra course... we have to explain every possible detail.

    Thank you for your help and insights!
  7. Oct 21, 2012 #6


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    Hm I see another way you could've done this.

    You know : [itex]\forall ε'>0, \exists N' \space | \space n>N' \Rightarrow |a_n - 0| < ε'[/itex]

    Simply because you are given that an → 0.

    You want : [itex]\forall ε>0, \exists N \space | \space n>N \Rightarrow |a_{n}^{2} - 0| < ε[/itex]

    So take what you want and massage it a bit :

    [itex]|a_{n}^{2} - 0| = |a_{n}^{2}| = a_{n}^{2} ≤ ε \Rightarrow a_n ≤ \sqrt{ε}[/itex]

    Now choosing [itex]ε' = \sqrt{ε}[/itex], what happens?
  8. Oct 23, 2012 #7
    It implies that [itex] a_n^2 < \varepsilon [/itex].

    So now I am trying to extend this to;
    Given [itex] \{a_n\}\rightarrow L[/itex] show that [itex] \{a_n^2\}\rightarrow L^2[/itex]. I have been playing around a bit and have come up with two ideas:

    Making the standard assumption:
    [tex] |a_n^2-L^2| = |a_n^2-a_nL-a_nL+L^2+a_nL-L^2+a_nL-L^2
    = |(a_n-L)^2 +2L(a_n-L)| < \varepsilon ^2 + 2L\varepsilon [/tex].

    I'm not 100% sure what can be done with that or that it implies anything based on the assumption that for n>N [itex]\{a_n\}\rightarrow L[/itex].

    My other idea is this:

    [tex] |a_n^2-L^2|=|a_n^2-a_nL+a_nL-L^2| \leq |a_n^2-a_nL|+|a_nL-L^2|

    If we now that [itex] \{a_n\}\rightarrow L[/itex] and set up our hypothesis so that [itex]|a_n-L|<\frac{\varepsilon}{2|L|} [/itex] can we extend the stuff immediately above to [tex] |a_n||a_n-L|+|L||a_n-L| < |a_n||\frac{ε}{2L}|+|L||\frac{ε}{2L}| < |L||\frac{ε}{2L}|+|L||\frac{ε}{2L}| < ε. [/tex]

    The above coming from the fact that the sequence is bounded by L.
  9. Oct 23, 2012 #8


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    Just like you wanted, right? So certainly we would have [itex]|a_{n}^{2} - 0| < ε[/itex] which would complete your proof.

    Most of these styles of proofs involve stating the definition you already know is true ( Usually because you are given one of your limits to be true and you can take it for granted ) and then proving the definition you want is also true by using the assumed definition.

    As for your second question :

    You know : [itex]\forall ε'>0, \exists N' \space | \space n>N' \Rightarrow |a_n - L| < ε'[/itex] ( This is your given limit you can take for granted ).

    You want : [itex]\forall ε>0, \exists N \space | \space n>N \Rightarrow |a_{n}^{2} - L^2| < ε[/itex] ( This is the limit you want to prove by using the limit you're taking for granted.

    I'll also tell you that your 'other idea' is the path you want to take.
  10. Oct 23, 2012 #9


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    No, this isn't necessarily true. [itex]a_n \rightarrow L[/itex] does not imply [itex]|a_n| \leq |L|[/itex]. Fortunately, you don't need [itex]|a_n|[/itex] to be bounded by [itex]|L|[/itex]. You just need it to be bounded by SOME bound, say [itex]M[/itex]. And this IS true: a convergent sequence is bounded. (Can you prove it?)
  11. Oct 23, 2012 #10
    Thanks for the replies! I've thought about them and this is what I have come up with:

    By hypothesis [itex]\{a_n\}\rightarrow L.[/itex] Since the limit of [itex]a_n[/itex] is L, we may chose [itex]N_2[/itex] such that [itex]\forall n>N_2 \Longrightarrow |a_n|<|L|+1[/itex]. Furthermore, given ε>0 we may choose [itex]N_1[/itex] such that

    [tex] \forall n>N_1 \Longrightarrow |a_n-L|< \frac{\varepsilon}{2|L|+1} .[/tex]

    Choose [itex]N=max(N_1,N_2)[/itex].

    [tex] \forall n>N \Longrightarrow |a_n^2-L^2|\leq |a_n-L|(|a_n|+|L|) \mbox{by triangle inequality} < \frac{\varepsilon}{2|L|+1}(|a_n|+|L|)< \frac{\varepsilon (2|L|+1)}{2|L|+1} = \varepsilon.[/tex]

    I appreciate your feedback! Thanks much.
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