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Thread: proof for sequences

  1. #1
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    proof for sequences

    Given that sequences a and b converges respectively, to the limits L1 and L2. Use def of limit of sequence to prove that if a>=b for all natural numbers n, then L1>=L2.

    Answer given is,
    Suppose on the contrary that L1 < L2. Let epsilon = (L2-L1)/2
    , which is positive.
    Let n = max(N1;N2) + 1. Then n > N1 and n > N2. So we have
    ...............

    a) why are they using epsilon = (L2-L1)/2
    b) what does it mean by 'n=max(N1,N2) + 1?' why using this?

    calculus is really giving me headache now.
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  2. #2
    ynj
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    Quote Originally Posted by serenalee View Post
    Given that sequences a and b converges respectively, to the limits L1 and L2. Use def of limit of sequence to prove that if a>=b for all natural numbers n, then L1>=L2.

    Answer given is,
    Suppose on the contrary that L1 < L2. Let epsilon = (L2-L1)/2
    , which is positive.
    Let n = max(N1;N2) + 1. Then n > N1 and n > N2. So we have
    ...............

    a) why are they using epsilon = (L2-L1)/2
    b) what does it mean by 'n=max(N1,N2) + 1?' why using this?

    calculus is really giving me headache now.
    The definition of the limit of sequence is
    $\displaystyle \forall\epsilon>0,\exists N,\forall n>N,|a_n-a|<\epsilon$,right?
    I guess you may have some trouble with understanding this definition. So you may define $\displaystyle \lim_{n\rightarrow \infty} a_n=a$as:
    if $\displaystyle n$ is large enough, then $\displaystyle |a_n-a|$will be as small as possible.
    Let's do this problem in this way.
    Let$\displaystyle \lim_{n\rightarrow \infty} a_n=L_1,\lim_{n\rightarrow \infty}b_n=L_2$
    Then we suppose $\displaystyle L_1<L_2$.
    Imagine a $\displaystyle N$ which is extremely large. This $\displaystyle N$ makes $\displaystyle |a_n-L_1|,|b_n-L_2|$ both extremely small from $\displaystyle n>N$. Since $\displaystyle L_1<L_2$,$\displaystyle |L_1-L_2|$can not be "extremely small". If you write $\displaystyle a_n\approx L_1,b_n\approx L_2$ for the sake of convenience, then we will have $\displaystyle a_n<b_n$since $\displaystyle a_n\approx L_1<L_2\approx b_n$. So we have found the contradiction.
    If you understand this, then all the things such as $\displaystyle (L_2-L_1)/2,\max\{N_1,N_2\}+1$will become technical ,trivial and easy to understand!!
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