Explore BrainMass

Explore BrainMass

    Converging Subsequences

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Theorem: Suppose that a sequence S of real numbers has a subsequence that converges to a real number a. Then S converges to a.

    I know this is true as an if and only if statement, but I need a counter example to show that just one converging subsequence isn't enough.

    Here are two sequences I'm considering: {1,-1,1,-1,1,-1...).
    Another is {0, 1,1/2, 1/3, ..., 1/n}.

    Our definition of a subsequence is: Let S={(n,an(subscript)}be a sequence of real numbers. A subsequence of S is a sequence T formed as follows:
    1) Let theta be a function with domain the counting numbers and range contained in the counting numbers, and such that theta(n)>theta(m) whenever n>m.
    2) T ={(m,a(theta(me)(subscript).
    To specify a subsequence, you must specify the function theta.

    Here's where I need help:
    1) How do I develop a subsequence from one of the sequences I have. (If you have a simpler sequence, please suggust using it). What is the domain, range, and the function theta?

    2) How do I show convergence? I know that I pick an r, but then what? I need to show how both S and T converge and that they don't converge to the same point.

    Thanks!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:00 pm ad1c9bdddf
    https://brainmass.com/math/basic-calculus/converging-subsequences-156007

    SOLUTION This solution is FREE courtesy of BrainMass!

    I will go ahead with your sequences.
    First, we should understand that subsequence is a part of the given sequence. For the sequence you have considered above, we can list out few subsequences as {1, 1, 1, 1, 1, .....} or {-1, -1, -1, -1, -1, .....} or {1, -1, 1, 1, -1, 1, 1, 1, -1, 1, 1, 1, 1, -1.....}. A subsequence must be defined on its own but its elements must be from the parent sequence considered.

    Here in the above example, the main sequence is {1,-1,1,-1,1,-1...}. This sequence moves between -1 and 1 hence is called Oscillating sequence and does not 'go' to any particular value or does not converge.
    Where as the subsequence {1, 1, 1, 1, 1, .....} converges to '1' as every element of it is 1.

    A sequence <aj> of real (or complex) numbers is said to converge to a real (or complex) number c if for every e > 0 there is an integer N > 0 such that if j > N then | aj - c | < e. The number c is called the limit of the sequence and we sometimes write aj --> c as n --> oo (infinity)
    More about the definition can be had at http://mathworld.wolfram.com/ConvergentSequence.html

    If a sequence does not converge, then we say that it diverges.
    In the second example you have considered, the sequence is convergent to zero. As n increases 1/n gets very close to zero. Hence we say it is convergent to zero. Every subsequence of it converges to zero. Hence the statement give above would have been true had it been stated as "Every subsequence of a convergent sequence is convergent"

    But if a sequence have a convergent subsequence, the sequence does not necessarily converge.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:00 pm ad1c9bdddf>
    https://brainmass.com/math/basic-calculus/converging-subsequences-156007

    ADVERTISEMENT