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Math Help - disprove a convergence question?

  1. #1
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    disprove a convergence question?

    i know that An->1

    i need to prove that (An)^n ->1

    but when i construct limit
    lim (An)^n
    n->+infinity

    i get 1^(+infinity) which says that there is no limit
    what do i do in this case in order to disprove that (An)^n->1

    ??
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  2. #2
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    It is true that \left( {A_n = \sqrt[n]{n}} \right) \to 1.
    But what about \left( {\sqrt[n]{n}} \right)^n \to ?
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  3. #3
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    where did you find the power of 1/n
    when i do the limit
    the base goes to 1 and the power goes to + infinity

    that is not solvable

    what to do??

    they question says prove/disprove

    how to disprove
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    i know that An->1

    i need to prove that (An)^n ->1

    but when i construct limit
    lim (An)^n
    n->+infinity

    i get 1^(+infinity) which says that there is no limit
    what do i do in this case in order to disprove that (An)^n->1

    ??
    Why not try this method? Your book ought to show that the Root and Ratio test always yield the same result. So

    \limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}

    And you have already asked this question already.

    Also let us try a three case scenario again

    Case #1: A_n\geqslant A_{n+1}\cdots

    It is clear then from the fact that A_n\to1 that there exists some N such that [tex]N\leqslant{n}[/math[ implies

    A_n\geqslant{1}

    From there it is clear then that

    1\leqslant{A_n}\leqslant{A_n^n}

    Or 1\leqslant\sqrt[n]{A_n}\leqslant{A_n}


    Case #2: A_n\leqslant{A_{n+1}}\cdots

    From here it is clear that there exists a N such that N\leqslant{n} implies 0\leqslant{A_n}\leqslant{1}

    And it should be clear then that A_n^n\leqslant{A_n}\leqslant{1}

    Or

    A_n\leqslant\sqrt[n]{A_n}\leqslant{1}


    Case #3: This is when A_n=A_{n+1}\cdots where the conclusion readily follows.
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  5. #5
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    how did you came to the conclusion that my limit equal this?

    <br /> <br />
\limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    how did you came to the conclusion that my limit equal this?

    <br /> <br />
\limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}
    Im sorry when I saw Plato's post I mistakenly believed you were asking to prove that A_n\to1\implies\sqrt[n]{A_n}\to{1}, forgive me.
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