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Math Help - Proof of Convergence

  1. #1
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    Proof of Convergence

    I'm Kinda Stuck With An Assignment:

    For y>1 Prove That \lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R} By Applying The Binomial Expansion of \left(1+x\right)^n, y=1+x

    Can Someone Please Figure That Out?
    Last edited by olekaiwalker; August 17th 2009 at 03:56 AM.
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  2. #2
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    Quote Originally Posted by olekaiwalker View Post
    For y>0 Prove That \lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R} By Applying The Binomial Expansion of \left(1+x\right)^n, y=1+x
    This not true as written. For example let y=\frac{1}{2}.

    Do you mean y>1?
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  3. #3
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    You're Right. It Should Be

    y>1
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  4. #4
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    Quote Originally Posted by olekaiwalker View Post
    I'm Kinda Stuck With An Assignment:

    For y>1 Prove That \lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R} By Applying The Binomial Expansion of \left(1+x\right)^n, y=1+x
    I was hoping someone else might help you. I hate this problem.
    Let say that n>2k~\&~k>\alpha
    n>n-1>n-2>\cdots>n-k+1>n-k>\frac{n}{2}
    x+1=y~\&~\left( {1 + x} \right)^n  > \frac{{n(n - 1) \cdots (n - k + 1)}}{{k!}}x^k  > \left( {\frac{n}{2}} \right)^k \frac{{x^k }}{{k!}}

    From that it follows 0 < \frac{{n^\alpha  }}{{\left( {1 + x} \right)^n }} < \frac{{k!}}{{x^k }}\frac{{2^k }}{{n^k }}n^\alpha   = \frac{{2^k k!}}{{x^k n^{k - \alpha } }}~~(k>\alpha).

    It should be clear that  \left( {\frac{{2^k k!}}{{x^k n^{k - \alpha } }}} \right) \to 0.
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