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Thread: limit

  1. #1
    Member disclaimer's Avatar
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    limit

    Hello;

    What is the limit of \frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})}  ^{\log{2}^{n+1}}} as n tends to infinity? To my mind, it's either \frac{1}{e} or 0. And I'm leaning towards \frac{1}{e}, but I'm not sure.

    Thank you very much.
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  2. #2
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    The limit is...

    I assume you are using base 10 in your logarithms. The limit is zero.
    This simplifies to:

     \left( {\frac {n}{n+1}} \right) ^{n\ln  \left( 2 \right) } \left( {<br />
\frac {1}{ \left( n+1 \right) \ln  \left( 2 \right) }} \right) ^{\ln <br />
 \left( 2 \right) }

    If you are using base e. A similar simplification will come out of base 10. So the limit is a little more obvious.
    Last edited by spammanon; Mar 7th 2008 at 07:05 PM. Reason: Try latex.
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  3. #3
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    Yeah..... I just found that out myself.... Thanks for your help anyway
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  4. #4
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    <br />
\frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})}  ^{\log{2}^{n+1}}}<br />
comes from the ratio test (all multiplied by 2) as an attempt to prove that <br />
\sum_{k=1}^{\infty}\frac{2^k}{({\log{2}^{k})}^{\lo  g{2}^{k}}}<br />
, which is derived from <br />
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}} (by the Cauchy condensation test), is convergent.

    By the way, what is the best way to show that <br />
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}} is convergent? Thanks.
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  5. #5
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    Quote Originally Posted by disclaimer View Post
    By the way, what is the best way to show that <br />
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}} is convergent? Thanks.
    How about using the extended version of the Cauchy condensation test, which says that you can replace the number 2 by any real number greater than 1?

    If you apply the condensation test with e in place of 2, it tells you that the convergence of \sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}} is equivalent to that of \sum\left(\frac en\right)^n, which converges by comparison with \sum\left(\frac 12\right)^n.

    Of course, if the logs are to base 10 then the same argument will work. You just replace 2 by 10 instead of e.
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