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Sequences, need help

  1. Jan 17, 2008 #1
    Sequences, need help!!

    Well there is this problem that i am struggling to proof, i think i am close but nope nothing yet. Well, the problem goes like this:
    We have two sequences [tex] \ {(a_n)} [/tex] and [tex] ({b_n})[/tex], whith the feature that [tex]{a_n}<{b_n}[/tex] for every n. Also, [tex]{a_2}={b_1}, {a_3}={b_2}... [/tex] and so forth. What i need to show is that they have the same limit, that is:

    [tex]\lim_{n\rightarrow infinit} {a_n}=m=\lim_{n\rightarrow infinit} {b_n}[/tex]

    I started by using the definition of the limit like this,
    First i assumed that the limits are not the same so i put:
    [tex]\lim_{n\rightarrow infinit} {a_n}=a[/tex] and [tex]\lim_{n\rightarrow infinit} {b_n}=b [/tex], where a is not equal to b, so it must be either smaller or bigger than it.
    For every epsilon, there exists some number N_1, that for every n>N it is valid that:
    I a_n-a I< epsylon, and also for every epsylon there must exist some number call it N_2, that for every n>N_2 we have:
    I b_n- b I< epsylon,

    then i took N=max{N_1,N_2} so for n>N, both
    I b_n- b I< epsylon and I a_n-a I< epsylon are valid

    then after some transformations i got:

    a- epsylon<a_n< a+ epsylon and
    b-epsylon< b_n < b+ epsylon

    let us now suppose that b<a, than this is not possible since from a theorem i could find a number call it n_o that for every n>n_o also b_n<b<a_n , which actually contradicts the very first supposition that b_n>a_n.

    now let us take b>a, and also from the supposition that b_n>a_n whe have

    a-epsylon<a_n<b_n<b+epsylon

    and this is where i get stuck.
    Any suggestions would be really appreciated.


    , so from here we have that the limit of a_n would be both b and
     
  2. jcsd
  3. Jan 17, 2008 #2
    dear suputidmath
    first observe that both a_n and b_n are strictly monoton increasing sequences. You can prove that every strictly monoton increasing sequence has a limit
    (there are two cases: the sequence is bounded or it is not).
    Take care that you cannot argue as you did if the limit a = infinity! infinity -epsilon < a_n makes no sense!
    When proving that a=b the case one is a real number and the other is infinity is simple.
    Now asume that both a and b are different real numbers.
    let d=abs(a-b).
    you know since a_n converges every epsilon neighbourhood of a contains almost all a_n (the same holds with respect to b and b_n). Have a look at the epsilon=d/3 and derive a contradiction.
     
  4. Jan 17, 2008 #3

    Gib Z

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    You could also use the fact that taking away a finite number of terms does not effect the limit of the sequence.
     
  5. Jan 17, 2008 #4
    Thankyou for your replys, but this very last part i cannot actually figure it out how does it help?? do i need to show for example that abs(a-b) should be contained to this epsylon interval of a, and then conclude that this is impossible since the interval is of a distance of 2epsylon, while the distance abs(a-b) is of a distance of 3epsylon.
    Can you please give me some more hints at this point??

    Anyones hints would be really appreciated!
    thnx
     
    Last edited: Jan 17, 2008
  6. Jan 17, 2008 #5

    Gib Z

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    Do you think my hints wrong or something >.<? Or you just want to do an epsilon delta proof? Or Perhaps this is for a proof of my hint and hence circular to use it?
     
  7. Jan 17, 2008 #6
    well actually i am looking for an epsylon delta proof. Using your hints, if i got it right, i could just take out the first term of the first sequence [tex] \ {(a_n)} [/tex]
    , in other words ommit the [tex] \ {a_1} [/tex] term so now the two sequences would have equal terms and so their limits would be the same. Because the first part i proved, the part to show that the limits of these strictly increasing sequences exist, in both cases, when they are upper bounded and not. If this is what you meant Gib Z, thanks anyways, but i am looking for an epsylon delta proof, because it is more rigorous, this what u said(and what i had in mind) is more like an intuitive aproach.
     
  8. Jan 17, 2008 #7

    Gib Z

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    But what I said in post 3 is a perfectly rigorous statement, a well known theorem. It can be proven.
     
  9. Jan 17, 2008 #8
    ok then, does the rest of reasoning follow like i said in post #6??
    And could you please post the whole theorem and the proof, this is the first time i hear it about sequences, i know that taking out or adding a finite number of terms to a series does not affect it, but as for sequences this is my first time i hear it, so i would really appreciate it Gib Z if you could post it, or just show me where i could find it.
    thnx.

    P.S. I really appreciate your help Gib Z, i really do and i am eager to see this kind of proof to my problem using your hints, but i also would like to know how to do it using delta epsylon.

    so????????
     
    Last edited: Jan 18, 2008
  10. Jan 18, 2008 #9
    an outline of your prove.
    a and b are different real numbers.
    d=abs(a-b) is a fixed real number depending on a and b

    [tex] \lim_{n \rightarrow \infty} a_n = a \in \mathbf{R}[/tex] means that [tex] \forall \epsilon >0 \exists N(\epsilon) [/tex] such that if [tex] n > N(\epsilon) \Rightarrow abs(a- a_n )< \epsilon [/tex].
    It must hold for all [tex] \epsilon > 0 [/tex] ,so you are free to chose [tex] \epsilon= \frac{d}{3} [/tex].
    let [tex] N_a=N(\epsilon)[/tex] for the series a_n and [tex] N_b=N(\epsilon) [/tex] for b_n.
    let [tex] M=\max (N_a,N_b)+10 [/tex]
    look at [tex] a_M = b_{M-1} [/tex]. use the triangle inequality with a,b,a_M
     
  11. Jan 20, 2008 #10

    Gib Z

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