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Thread: Logarithmic limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Logarithmic limit?

    Evaluate:

    $\displaystyle \lim_{n\rightarrow \infty} \{\log_{n - 1} n\cdot \log_{n} (n + 1)\cdot \log_{n + 1} (n + 2)\mbox{...}\log_{n^k - 1} n^k\}$, where $\displaystyle k\in\mathbb{N}$
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  2. #2
    Senior Member Spec's Avatar
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    $\displaystyle \log_{n-1}(n)\cdot\log_{n}(n+1)\cdot\log_{n+1}(n+2)\cdot.. .\cdot\log_{n^k-1}(n^k)=$ $\displaystyle \frac{\ln(n)}{\ln(n-1)}\frac{\ln(n+1)}{\ln(n)}\frac{\ln(n+2)}{\ln(n+1) }\cdot ... \cdot \frac{\ln(n^k)}{\ln(n^k-1)}=$ $\displaystyle \frac{\ln(n^k)}{\ln(n-1)}$

    Using L'Hospital's rule I get the answer to be $\displaystyle k$.
    Last edited by Spec; Jun 9th 2009 at 12:47 PM.
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  3. #3
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    Brilliant work, spec!


    I'd finish it this way . . .

    $\displaystyle \lim_{n\to\infty}\frac{\log(n^k)}{\log(n-1)} \:=\:\lim_{n\to\infty}\frac{k\!\cdot\!\log(n)}{\lo g(n-1)}\;=\;k\cdot\lim_{n\to\infty}\frac{\log(n)}{\log (n-1)} \;=\;k\cdot1 \:=\:k$

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