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Prove that a convergent sequence is bounded

  1. Aug 20, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and solution are attached as TheProblemAndSolution.jpg.

    2. Relevant equations
    Definition of the limit of a sequence.

    3. The attempt at a solution
    I understand how P = ϵ + |A| can be seen as an upper bound that proves that the sequence is bounded, but for the last bit of the solution where the solution says “we choose P as the largest one of the numbers [. . .]”, why does it say a_N instead of a_n, shortly after? Is that a typo? Also, there isn't necessarily a value of a_n that is the largest (such that no value of a_n will ever reach ϵ + |A|), unless the author(s) meant to say that ϵ + |A| is one of the numbers to choose from such that ϵ + |A| is the largest number mentioned by the solution.

    Could someone please clarify the last part of the solution for me?

    I know this is a pedantic request, but I would greatly appreciate any input!
     

    Attached Files:

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  3. Aug 20, 2014 #2

    Zondrina

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    No that is not a typo.

    ##|a_n| < \epsilon + |A| = P## for ##n > N##.

    By hypothesis you found an ##N## such that for any ##n ≥ N, \space |a_n - A| < \epsilon##.

    This means you can bound ##|a_n|## within ##P## for a sufficiently large ##n##. This ##P## corresponds to the largest of the numbers ##a_1, ..., a_N, (\epsilon + |A|)## since we already found the ##N## such that ##|a_n - A| < \epsilon##.
     
  4. Aug 20, 2014 #3

    s3a

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    Oh, I think I get it now!

    [1] I thought ϵ + |A| would always be the largest number, but if a_n = 1/n, for example, then a_1 would be the largest of the numbers at the end of the solution, right? Furthermore, it will always either be a_1 or ϵ + |A| that would be the largest number, and not a_2, . . ., a_(N–1), right?

    [2] Though, since we're using n ≥ N (along with |a_n – A| < ϵ|) to define the limit (as can be seen by the fact that the numbers range from a_1 to a_N instead of a_1 to a_(N–1), at the end of the solution), that also means that |a_n| < ϵ + |A| for n ≥ N (instead of n > N), right?

    I know this last minor detail makes no difference in the grand scheme of things, but I am a pedantic person. :P
     
  5. Aug 20, 2014 #4

    s3a

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    Also, when the author(s) said to choose the largest one of the numbers for P, "largest", in this context, means with greatest magnitude, such that –1 is "larger" (in magnitude/absolute value) than +1/2, right?
     
  6. Aug 20, 2014 #5

    PeroK

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    Hi s3a. That solution you posted is not very good. You've identified one problem. It really should be the largest of |a_1| ... |a_N|.

    But, also, you're looking for P > |a_n|, so really it should be the largest of |a_1| ... |a_N| + 1.

    Third, although it's sort of okay to leave ε as an arbitrary +ve number, it would seem better to me to take ε = 1, say, and find the value of N for which n > N => |a_n - A| < 1.

    So, P = max{|a_1| ... |a_N|, |A|} + 1
     
    Last edited: Aug 20, 2014
  7. Aug 20, 2014 #6

    PeroK

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    PS Don't forget to look for examples and not just rely on the analysis. E.g:

    a_1 = 3, a_2 = 2, a_3 = 1, a_4 = 1/2, a_5 = 1/4 ...

    In this case A = 0 and when ε = 1, we have N = 3:

    n > 3 => |a_n - 0| < 1 => |a_n| < 1

    P = max{|a_1|, |a_2|, |a_3|, |A|} + 1 = max{3, 2, 1, 0} + 1 = 4
     
  8. Aug 21, 2014 #7

    s3a

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    Thanks for your answer, Perok.

    Okay, so it seems that you're saying that P = max{|a_1|, |a_2|, . . ., |a_N|, |A|} + ϵ.

    However, it seems to me that |a_1|, |a_2|, . . ., |a_N| each already have an error margin that is greater than or equal to ϵ, with respect to |A| (instead of strictly less than ϵ, because, for the error margin to be strictly less than ϵ, n must be greater than N), so, from that, it seems to me like it should be P = max{|a_1|, |a_2|, . . ., |a_N|}, since it seems to me that |a_i| ≥ |A| + ϵ, where i is any number from [1, N] (or [0, N], if a_0 is defined).

    If I'm wrong, could you please give me an example where |a_i| < |A| + ϵ?
     
  9. Aug 21, 2014 #8

    PeroK

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    a_1 = 0
     
  10. Aug 21, 2014 #9

    s3a

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    Thanks for that.

    What about a sequence formula, where there is at least one nonzero a_i such that |a_i| < |A| + ϵ?

    Would you be able to give me such an example, by any chance? :)
     
  11. Aug 21, 2014 #10

    PeroK

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    ##a_n = 1 - (\frac{1}{2})^n##

    Don't get hung up on formulas. The mathematical properties of a thing do not depend on expressing that thing as a formula.
     
  12. Aug 21, 2014 #11

    LCKurtz

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    Would you be able to come up with an example yourself? Have you tried?
     
  13. Aug 23, 2014 #12

    s3a

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    Thanks again, Perok.

    Okay, so for n ≤ N, a_n = 1 – (1/2)^n seems to be a formula for which |a_i| < |A| + ϵ.

    It also seems that, for n ≤ N, a_n = (1/2)^n is a formula for which |a_i| ≥ |A| + ϵ.

    About the formulas, I just needed the formulas to build my intuition.

    And, about finding one myself, I think I was a little too confused with everything at the time.

    I see now that the a_1, a_2, . . ., a_N shouldn't be a_1, a_2, . . ., a_n, because for n > N, it would always be the case that |a_n| < |A| + ϵ and |A| + ϵ is one of the candidates for the value of P, and it's pointless to add candidates for P that are guaranteed to not be the largest.

    I have one last confirmatory question, shown below.:
    Since there are two definitions for the set of natural numbers (one including 0 and another excluding 0), if we used the definition of natural numbers including 0, we would have a_0, a_1, a_2, . . ., a_N instead of a_1, a_2, . . ., a_N as the candidates for P (in addition to |A| + ϵ), right?
     
  14. Aug 23, 2014 #13

    Fredrik

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    Since a complete solution has already been given, I will offer my version:

    The key to this problem is the realization that "almost all" terms are close to A, which is equal to |A| or -|A|, depending on whether A is negative or non-negative.

    Suppose that ##\lim_n a_n =A##. Then every open interval with A at the center contains all but a finite number of the ##a_n##. In particular, the open interval ##(A-1,A+1)## contains all but a finite number of the ##a_n##. We don't know if A is negative or non-negative, but in both cases, the open interval ##(A-1,A+1)## is a subset of the closed interval ##[-|A|-1,|A|+1]##. Let N be a positive integer such that ##a_n\in [-|A|-1,|A|+1]## for all ##n\geq N##. Some of the first N-1 terms may be outside of the interval [-|A|-1,|A|+1], so we can't just set P equal to |A|+1. But if we set ##P=\max\{|a_1|,\dots,|a_{N-1}|,|A|+1\}##, the interval [-P,P] will contain all terms.
     
    Last edited: Aug 23, 2014
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