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Math Help - proof of limit

  1. #1
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    proof of limit

    Hi:

    how could one prove that if a_n --> A and A != 0 then

    a_(n+1)/a_n --> 1 ?

    I think some average inequalities could help here but I can't see it.

    Thanx
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tonio View Post
    Hi:

    how could one prove that if a_n --> A and A != 0 then

    a_(n+1)/a_n --> 1 ?

    I think some average inequalities could help here but I can't see it.

    Thanx
    For large enough n,

    |a_{n+1}-A|<\epsilon\implies\left|\frac{a_{n+1}}{a_n}-\frac{A}{a_n}\right|<\frac{\epsilon}{a_n}

    Taking n\to\infty (and noting that a_n\to A) gives us:

    \left|\lim_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\r  ight)-\frac{A}{A}\right|<\frac{\epsilon}{A}\implies\left  |\lim_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\right)-1\right|<\frac{\epsilon}{A}

    which proves the theorem.

    An easier, though less technical way of doing is noting that

    \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{\displa  ystyle\lim_{n\to\infty}a_{n+1}}{\displaystyle\lim_  {n\to\infty}a_n}=\frac{A}{A}=1
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  3. #3
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    Huge. thanx a bunch, Redsoxfan
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