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Math Help - why this expression is true..

  1. #1
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    why this expression is true..

    <br />
\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}<br />

    ln is not always bigger then 1
    so when i am doing the comparing test
    i cant use that
    because ln 1 =0

    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    <br />
\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}<br />

    ln is not always bigger then 1
    so when i am doing the comparing test
    i cant use that
    because ln 1 =0

    ??
    because \ln{n} > 1 for n \geq 3
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    <br />
\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}<br />

    ln is not always bigger then 1
    so when i am doing the comparing test
    i cant use that
    because ln 1 =0

    ??
    But <br />
\sum_{n=3}^{\infty}\frac{\ln n}{n}> \sum_{n=3}^{\infty}\frac{1}{n}<br />

    is true so you can use the comparision test on that series and add the first couple of terms. Adding a few terms to the series won't change its convergenece (divergence).
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  4. #4
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    Nevermind that.

    ln(n)/n may initally be smaller, but when n > e it overtakes 1/n.

    Seeing as both terms are over some number n, the determining factor is the size of the numerator. 1 will always be 1, but ln(n) is always increasing.
    Last edited by derfleurer; May 3rd 2009 at 02:59 PM. Reason: more or less insignificant
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  5. #5
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    More accurately, ln n is bigger than 1 when n is bigger than e (or 2.718)
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