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Math Help - Easy proof - Logarithms

  1. #1
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    Easy proof - Logarithms

    Well, my problem is this : I need to check if this series is convergent :

    Sigma of (1 to infinity) : \frac{log(n)}{2n^3-1}

    now, I did the following thing:

    \frac{log(n)}{2n^3-1} < \frac{log(n)}{2n^3} < \frac{2n}{2n^3}= \frac{1}{n^2}

    *Wherever I wrote ' < ', it also works with ' <= '
    (The last one is known to be convergent)

    All is left for me to prove is this:

    2n>=log(n)

    I tried to prove it with induction... We haven't yet 'officially' learned logarithms, so this is not obvious to use.

    Thanks
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  2. #2
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    Quote Originally Posted by adam63 View Post
    Well, my problem is this : I need to check if this series is convergent :

    Sigma of (1 to infinity) : \frac{log(n)}{2n^3-1}

    now, I did the following thing:

    \frac{log(n)}{2n^3-1} < \frac{log(n)}{2n^3} < \frac{2n}{2n^3}= \frac{1}{n^2}

    *Wherever I wrote ' < ', it also works with ' <= '
    (The last one is known to be convergent)

    All is left for me to prove is this:

    2n>=log(n)

    I tried to prove it with induction... We haven't yet 'officially' learned logarithms, so this is not obvious to use.

    Thanks
    You could consider the function

     <br />
f(x) = 2x - \ln x<br />
    and show that f(1) > 0 and f'(x) > 0 for x > 1 so f(x) > 0 for all x > 1 (MVT).
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