Prove the limit of this sequence is 1

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October 26th 2010, 01:22 AM
Newtonian

Prove the limit of this sequence is 1

Prove that

$\displaystyle\lim_{x\to\infty} \frac{2}{(\ln(x))^2} \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} = 1$
October 26th 2010, 11:39 AM
zzzoak

As x goes to infinity the sum

$ \displaystyle { \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} } $

may be evaluated

$ \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 } $

$ 2 \; \displaystyle { \int_{1}^{x} \frac{ln \; t}{t+1}dt= ln^2x $
October 26th 2010, 01:26 PM
Newtonian

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?
October 26th 2010, 03:21 PM
zzzoak

This sum is a Riemann sum of this integral.

For x going to infinity

$ ln(1+x)=ln(x(1+1/x))=lnx+ln(1+1/x)=lnx+1/x +o(1/x) $
October 30th 2010, 08:19 AM
Newtonian

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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