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Math Help - Prove the limit of this sequence is 1

  1. #1
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    Prove the limit of this sequence is 1

    Prove that

    \displaystyle\lim_{x\to\infty}  \frac{2}{(\ln(x))^2} \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} = 1
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  2. #2
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    As x goes to infinity the sum

    <br />
\displaystyle { \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} }<br />

    may be evaluated

    \; =ln \; t \:\; ln(t+1) \; |_1^x - \int_{1}^{x} \frac{ln \; (t+1)}{t}dt
    }
    " alt="
    \displaystyle { \int_{1}^{x} \frac{ln \; t}{t+1}dt= \int_{1}^{x} ln \; t \:\; d( \; ln(t+1) \ \; =ln \; t \:\; ln(t+1) \; |_1^x - \int_{1}^{x} \frac{ln \; (t+1)}{t}dt
    }
    " />

    <br />
2 \;   \displaystyle { \int_{1}^{x} \frac{ln \; t}{t+1}dt=  ln^2x<br />
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  3. #3
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    Sorry, I don't follow what you're trying to say. The sum may be approximated by the integral, but they're not equal, and the integral isn't equal to (\ln(x))^2 so I'm not quite sure what you're trying to show...could you explain a bit more please?
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  4. #4
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    This sum is a Riemann sum of this integral.

    For x going to infinity

    <br />
ln(1+x)=ln(x(1+1/x))=lnx+ln(1+1/x)=lnx+1/x +o(1/x)<br />
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  5. #5
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    OK I see what you're saying now. I think everything's fine as long as we can show that the approximation of the sum by the integral at the beginning is valid; that is, we need to show

    \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} \sim \int_{1}^{x} \frac{\ln(t)}{t+1}

    as x \rightarrow \infty, where f(x) \sim g(x) means f(x)/g(x) \rightarrow 1 as x \rightarrow \infty. I've thought about this for a while; is it just a standard kind of integration result or can you prove it?
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