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Thread: Equivalence of definition (limited sequence)

  1. #1
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    Equivalence of definition (limited sequence)

    Hello,

    I am asked to show that the following two definitions of the limit $\displaystyle L$ of a sequence $\displaystyle s_{n}$ are equivalent:

    a) $\displaystyle \forall \epsilon>0 \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < \epsilon$
    b) $\displaystyle \forall m \in \mathbb{Z}_{+} \exists N\in \mathbb{R} \forall n \geq N: abs(s_{n}-L) < \frac{1}{m}$

    I do see it intuitively, but how do I show it quantitatively?

    Thanks.
    Last edited by detalosi; Jul 30th 2019 at 09:09 AM.
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  2. #2
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    Re: Equivalence of definition (limited sequence)

    $\text{show that $a \Leftrightarrow b$}$
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  3. #3
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    Re: Equivalence of definition (limited sequence)

    So I must show that:

    1) $\displaystyle a \Rightarrow b$
    2) $\displaystyle b \Rightarrow a$

    I start with 1):

    I assume that a sequence $\displaystyle \{S_{n}\}$ converges to $\displaystyle L$ according to 1).
    This means that for every $\displaystyle \epsilon>0$ I can find an $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \mid S_{n}-L \mid < \epsilon$ whenever $\displaystyle n\geq N$.
    Since there is no restriction on $\displaystyle \epsilon$ other than that it has to be positive I let $\displaystyle \epsilon=m$ where $\displaystyle m\in \mathbb{Z_{+}}$. Not sure!

    Hints would be appreciated!
    Last edited by detalosi; Jul 31st 2019 at 04:03 AM.
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  4. #4
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    Re: Equivalence of definition (limited sequence)

    if we let $\displaystyle \epsilon=m$ then $\displaystyle a)$ says

    $\displaystyle \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < m$

    whereas you are trying to show $\displaystyle < \frac{1}{m}$
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  5. #5
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    Re: Equivalence of definition (limited sequence)

    Quote Originally Posted by detalosi View Post
    Hello,

    I am asked to show that the following two definitions of the limit $\displaystyle L$ of a sequence $\displaystyle s_{n}$ are equivalent:

    a) $\displaystyle \forall \epsilon>0 \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < \varepsilon$
    b) $\displaystyle \forall m \in \mathbb{Z}_{+} \exists N\in \mathbb{R} \forall n \geq N: abs(s_{n}-L) < \frac{1}{m}$
    I think that you have a typo in part b). Is it not $\Large\exists N\in \mathbb{Z}~?$

    So to prove that $a \Rightarrow b$ start with $m \in \mathbb{Z}^{+}$
    Using a); suppose that $0<\varepsilon<\frac{1}{m}$ we get $\exists N\in\mathbb{Z}^+$ such that if $n\ge N$ then $|L-s_n|<\varepsilon<\frac{1}{m}$

    To prove that $b \Rightarrow a$ start with $\varepsilon>0$.
    Using b); we know that $\exists N\in\mathbb{Z}$ having the property that if $n\ge N$ then $|L-s_n|<\varepsilon$
    By simple algebra we know that if $n>N $ then $\frac{1}{n}<\frac{1}{N}<\varepsilon$.
    Thus if $n\ge N$ then $|L-s_n|<\varepsilon$
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  6. #6
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    Re: Equivalence of definition (limited sequence)

    Thanks.
    The definition states "... for every positive integer m there is a real number N so that ...". I am afraid it is not a typo.
    Last edited by detalosi; Aug 2nd 2019 at 01:16 PM.
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