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A small question about Cauchy sequence

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Let a_n = (1/2)[(1/a_n)+1] and a_1=1, does this sequence converge?

    2. Relevant equations

    A sequence in R^n is convergernt if and only if it's cauchy.
    A sequence in R^n is called a cauchy sequence if x_k - x_j ->0 as k, j-> infinity.

    3. The attempt at a solution

    I am confused about what the solution says below:

    a_2 = 1, so all points a_i =1. therefore the sequence is equal to 1,1,1,1 ...
    (so far I am ok)
    this sequence converges to 1 by the cauchy condition: a_n+1 - a_n = 0 which goes to 0 as n goes to infinity. since limit points are unique, the sequence converges to 1.

    But as the definition of cauchy sequence says, it tends to 0 but should not equal to 0 as k,j goes to infinity.

    but here, all the elements are equal to 1, so x_k - x_j = 1 for all k,j. That's not the condition of limit, is it? cuz limit shouldn't be "equal", limit just can be tended to, right?
     
  2. jcsd
  3. Oct 4, 2008 #2

    statdad

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    Homework Helper

    Constant sequences (all terms equal) or sequences that become constant after a finite number of terms do converge.

    A sequence [tex] \{a_n\} [/tex] converges to a number [tex] a [/tex] if, for any [tex] \varepsilon > 0 [/tex] you can find an integer [tex] N [/tex] so large that

    [tex]
    |a_n - a| < \varepsilon \quad \text{ for all }n \ge N
    [/tex]

    That is certainly true for your sequence.

    Your sequence is also Cauchy: if all the terms equal one, then

    [tex]
    |a_n - a_m| = 0 < \varepsilon
    [/tex]

    no matter how small [tex] \varepsilon [/tex] might be.
     
  4. Oct 4, 2008 #3
    I get it , thanks statdad.

    but I have a question on your explanation above., how would you explain the second sentence on the graph.
    i mean , like the proof of least upper bd M, x > M - epislon for any epislon>0
    in this epislon proof, i understand the meaning , which is, no matter how small epislon is, (once it's a little bit smaller than M, then there exist an element x in the set S to substitute the original least upper bd M.)

    but can you explain the meaning of the definition you stated?...cuz i really dont get it much. thx thx
     
  5. Oct 4, 2008 #4
    Can you give the precise meaning of "x_k - x_j ->0 as k, j-> infinity" I think it will help.
    Also note that checking if a_n+1 - a_n goes to 0 as n goes to infinity is totally different from "if x_k - x_j ->0 as k, j-> infinity"
     
  6. Oct 4, 2008 #5
    why totally different??
    i thought they were in the same pattern


    x_k-x_j->0 as k,j -> infinity, that is if for every epislon>0 there exists an integer K s.t. /x_k - x_j/<epislon whenever k>K and j >K
     
  7. Oct 4, 2008 #6
    let S_n= 1+1/2+1/3+...+1/n , and you will know that S_n is not cauchy and hence diverges in any typical analysis books.
    Why it's not cauchy? because it contradict to "x_k-x_j->0 as k,j -> infinity" (you may state it precisely and try to prove it )
    But, you will find that S_n - S_n-1 = 1/n, which definitely goes to 0, as n goes to infinity.

    "/x_k - x_j/<epislon whenever k>K and j >K" is right and note that it's different from "0</x_k - x_j/<epislon" hence x_k=x_j is allowed in this case. You will perhaps find "0</x - y/<epislon" is required somewhere else.
     
    Last edited: Oct 4, 2008
  8. Oct 4, 2008 #7

    statdad

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    pantin;
    I'm not sure to which 'graph' you mean - equation?

    The first one I presented was this:

    The sequence [tex] \{a_n\} [/tex] converges to [tex] a [/tex] if, given any [tex] \varepsilon > 0 [/tex], there is an integer [tex] N [/tex] such that
    [tex]
    |a_n - a| < \varepsilon \quad \text{ for } n > N
    [/tex]

    What does this mean? An incredibly non-mathematical, but rather intuitive, description of a sequence converging to a number is that eventually (meaning when [tex] n [/tex] is 'large enough'), all the terms of the sequence will be 'very close' to that number.

    What is the [tex] varepsilon [/tex] in the above formula? It answers the question "how close to the proposed limit do we want the terms?"

    What is the purpose of the [tex] N [/tex] in the formula? It tells us how many terms into the sequence we need to move in order to find terms close to a.

    Regarding the notion for a Cauchy sequence: again, intuitively, a sequence is a Cauchy sequence if eventually its terms are all "very close to each other".

    The notation

    [tex]
    |a_n - a_m |
    [/tex]

    is simply the distance between two arbitrary terms of the sequence. Mathematically, a sequence is a Cauchy sequence if, given any [tex] \varepsilon > 0 [/tex]
    we can find an integer [tex] N [/tex] such that

    [tex]
    |a_n - a_m| < \varepsilon \quad \text{for any choices of } n, m \text{ both } \ge N
    [/tex]

    A final comment - another post has popped up.
    You wondered why
    [tex]
    |a_n-a_{n+1}| < \varepsilon
    [/tex]
    for any choice of [tex] n [/tex]
    is different from
    [tex]
    |a_n - a_m | < \varepsilon
    [/tex]
    for any choice of [tex] n, m[/tex]. Two reasons.
    1. You statement means that you know any two consecutive terms are close to each other
      [*]The other notation means that any two terms, consecutive or widely different in position, are close to each other


    Finally, since to be Cauchy it has to be the case that no matter how small a number we choose for [tex] \varepsilon [/tex], that

    [tex]
    |a_n - a_m | < \varepsilon
    [/tex]

    must be true, the smaller the value of [tex] \varepsilon [/tex] the closer the difference is to zero, so we write

    [tex]
    \lim_{\{n,m\} \to \infty} |a_n - a_m| \to 0
    [/tex]
     
  9. Oct 4, 2008 #8
    very clear!
    (meaning when is 'large enough'), all the terms of the sequence will be 'very close' to that number.


    that's the explanation I want!
    thx a lot statdad! thx boombaby!
     
  10. Oct 4, 2008 #9
    so /a_n+1 - a_n/ < epislon is one of the case of /a_k - a_j/ < epislon right. but it cannot stand for all the cases
     
  11. Oct 4, 2008 #10

    statdad

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    "so /a_n+1 - a_n/ < epislon is one of the case of /a_k - a_j/ < epislon right. but it cannot stand for all the cases"

    exactly. it is the case where you are considering only consecutive terms in the sequence.
     
  12. Oct 5, 2008 #11

    HallsofIvy

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    Staff Emeritus
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    Good heavens, no! If a_i= 1, for i> 1, as you say, then x_k- x_j= 1- 1= 0, not 1!
     
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