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Limit of a sequence

  1. Nov 4, 2004 #1
    Hi!

    Have I understood the concept of the limit of a sequence correctly?

    Let's say we have a sequence {x_n} (dots for spacing):

    ----|------------------|-------------------|----------|------->
    a-epsilon...................a.....................a+epsilon.........R............x_n

    So "a" is the limit of {x_n} if there exists an n_0 = f(epsilon) so that for every n > n_0 the inequality |x_n - a| < epsilon is true.

    How can |x_n - a| < epsilon be true for values of n which go beyond a+epsilon? For example, for R the inequality |R - a| < epsilon is not true, yet the definition of a limit of a sequence says that the inequality must hold for all n > n_0 (including that n which gives us R).

    I guess they avoid that problem on the "left side" of "a" by saying that n_0 is a function of epsilon and that automatically cuts off all values of "n" before a-epsilon. Why is that not done on the "right side"? They allow "n" to increase towards positive infinity so that the inequality won't hold past the value of "n" which gives us a+epsilon.

    - Kamataat
     
  2. jcsd
  3. Nov 4, 2004 #2

    arildno

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    No, no!!
    You must have:
    [tex]|x_{R}-a|<\epsilon[/tex]
    [tex]x_{R}[/tex] represents the R'th term in the sequence, and there is no reason why this term should equal R.
     
  4. Nov 4, 2004 #3
    Yes, I know.

    In my inequality |x_n - a| < epslion the "x_n" term means n-th element of the sequence.

    Why can't it equal R? The definition says that for all n > n_0. If n_0-th element of the sequence gives us the value a-epsilon, then if we start increasing n, we can get values between a-epsilon and a, a and a+epsilon and values greater than a+epsilon. But for values greater than epsilon, the inequality is no longer true. But "a" to be the limit, the inequality has to be true for all n > n_0 including those which give us values greater than a+epsilon. But it can't be true for values greater than a+epsilon.

    - Kamataat
     
    Last edited: Nov 4, 2004
  5. Nov 4, 2004 #4

    matt grime

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    But, there need not be any particular n such that x_n=R, indeed by definition, if x_n is convergent etc.. there is no n greater than n_0 such that x_n=R.

    I don't think you understand what a sequence is from the looks of it.


    A sequence is a set of elements of, in this case, the real numbers indexed by the natural numbers, eg

    x_n = 1/n,

    or
    x_n = 2 for all n


    Both of those converge, the first to zero the second to 2.



    It doesn't make sense to say "those values of n which go beyond a+e", at least not to me: what goes beyond a+e? x_n? why must it?
     
  6. Nov 4, 2004 #5
    I do understand what a sequence is.

    When I say "those values of n which go beyond a+e" I mean values of n which give us elements of the sequence with values greater than a+e.

    Grrrrrrrrr. My bad. I understand it now (I hope). For some reason I was thinking that a can be in the middle of the sequence. GRRRRRRRRRRRRRRRRRRRRRRRRRRRR!!!!!

    So if the sequence converges to a then there is no n > n_0 that gives us values of the sequence's elements greater than a (or on the right side of a)?

    - Kamataat
     
    Last edited: Nov 4, 2004
  7. Nov 4, 2004 #6

    matt grime

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    No, if x_n converges to a, and we've chosen n_0 given that e, then there is no x_n>a+e for n>n_0

    1/n converges to zero, yet all terms in the sequence are greater than 0.
     
  8. Nov 4, 2004 #7

    shmoe

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    That's not correct. There is no n>n_0 where x_n is greater than a+epsilon, x_n could still be greater than a.
     
  9. Nov 4, 2004 #8
    How can x_n be greater than a? Doesn't the fact that a is the limit of x_n mean that x_n converges to a?
     
  10. Nov 4, 2004 #9

    matt grime

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    You agree that 1/n converges to zero? And what's special about greater than? why not less than too, since it is just a mirror image, and thus you're effectively saying that only constant sequences converge. Doesn't the fact that your objection has no mention of epsilon worry you?
     
  11. Nov 4, 2004 #10
    Yes, I agree.

    I was talking about greater than beacuse I was assuming an increasing sequence.

    Maybe I should ask differently.

    If we approach the limit from one side, then can there be any values of the sequence on the other side of that limit? So for an increasing sequence for which a=5, can there be values of x_n > 5? As I understand it, the answer is no.

    - Kamataat
     
  12. Nov 4, 2004 #11

    matt grime

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    Why didn't you say the sequence was increasing?

    If x_n is a strictly increasing sequence then obviously the limit must be greater than any of the elements in the sequence, if the limit exists.
     
  13. Nov 4, 2004 #12
    I thought you guys would get it out of the context, but I'm sorry if I wasn't clear enough.

    At last I get it. It seems obvious to me too.

    However, now I have another question. Why e>0? For an increasing sequence, that starts with negative values, passes 0 and converges to a positive value, couldn't I choose a negative e? I mean, does x_n0 = e?

    - Kamataat
     
    Last edited: Nov 4, 2004
  14. Nov 4, 2004 #13

    matt grime

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    Because the correct statement is

    |x_n-a|<e

    right?

    Now abs values are always positive, so if you picked e negative, then it would make no sense.
     
  15. Nov 4, 2004 #14
    I edited my post, but in case you didn't notice (I edited it shortly before you replied) I'll ask again:

    Does x_n0 = e or does x_n0 = a+/-e (depending on whether the sequence is increasing or decreasing)?

    - Kamataat
     
  16. Nov 4, 2004 #15

    matt grime

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    What is x_n0? My post still is the answer to why we have e positive: limits use abs values, if e were negative in that definition, we would asking for something that must be positive to be strictly negative, and that is contradictory. Again, look back at where the e, for epsilon, arises.
     
  17. Nov 4, 2004 #16
    x_n0 is the n_0-th term of the sequence.

    So basically I'm asking that if I draw the sequence on an axis then where on that axis would e be?

    - Kamataat
     
  18. Nov 4, 2004 #17

    matt grime

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    How do you know x_n0 =e? At what point do you see equality in any of the parts of the definition of limit? e does not "live in the same space" as the elements in the sequence.

    x_n tends to a if for all e>0, there is an n_0 such that |x_n-a| < e for all n> n_0

    at no point am I requiring x_n_0 to equal e, or to equal a+/-e, I'm not even declaring that a+e even makes sense, only that there is a distance notion in the space where we're talking about taking limits.

    It is only in the special case of Real numbers that we can talk about a+/-e.

    Where e is on the real axis is entirely immaterial to this argument.


    So, 1/n tends to zero, obviousl, so Iknow that for e=1/sqrt(2) etc.. even though there is no element in the sequence with n_0= 1/sqrt(2).
     
  19. Nov 4, 2004 #18
    1.) Well, since I am currently only dealing with real numbers, then can I assume a+/-e?

    2.)
    If e can be outside of the "space of the sequence" then what need is there for n_0 to be dependent on e? Couldn't I just as well say that "x_n tends to a if there is an n_0 such that |x_n-a| < e for all n> n_0"?
    - Kamataat
     
    Last edited: Nov 4, 2004
  20. Nov 4, 2004 #19

    matt grime

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    You can assume a+/e is what, does what, erm what?
     
  21. Nov 4, 2004 #20

    matt grime

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    You could say that. However almost always the n_0 will be dependent on e.

    Have you considered thinking about what's going on with some simple examples?
     
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