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Math Help - limit and supremum

  1. #1
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    limit and supremum

    Ok, this might be a silly/easy question but I have to make sure. Let say we have a seq of pos. numbers \{x_n\} such that x_n<x_{n+1} \forall n and lim_{n\rightarrow \infty}(x_n/x_{n+1})=1. Can I then, knowing the above, conclude that \sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1?
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    Re: limit and supremum

    Quote Originally Posted by Zeke View Post
    Ok, this might be a silly/easy question but I have to make sure. Let say we have a seq of pos. numbers \{x_n\} such that x_n<x_{n+1} \forall n and lim_{n\rightarrow \infty}(x_n/x_{n+1})=1. Can I then, knowing the above, conclude that \sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1?
    If 0<a<b then is it true that \frac{a}{b}<1~?
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  3. #3
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    Re: limit and supremum

    Yes, ofcourse it's true.
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  4. #4
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    Re: limit and supremum

    Quote Originally Posted by Zeke View Post
    Yes, ofcourse it's true.
    Well then each term of \left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\} < 1.
    Thus \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right) \le 1.

    What is the contradiction if we suppose that \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<1~?
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    Re: limit and supremum

    Quote Originally Posted by Plato View Post
    Well then each term of \left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\} < 1.
    Thus \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right) \le 1.

    What is the contradiction if we suppose that \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<1~?

    \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<\frac{a_{n}}{a_{n+1}}

    For some n
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