Prove that
$\displaystyle \displaystyle\lim_{x\to\infty} \frac{2}{(\ln(x))^2} \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} = 1$
As x goes to infinity the sum
$\displaystyle
\displaystyle { \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} }
$
may be evaluated
$\displaystyle
\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
}
$
$\displaystyle
2 \; \displaystyle { \int_{1}^{x} \frac{ln \; t}{t+1}dt= ln^2x
$
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 $\displaystyle (\ln(x))^2$ so I'm not quite sure what you're trying to show...could you explain a bit more please?
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
$\displaystyle \sum\limits_{1 \leq k \leq x-1} \frac{\ln k}{k+1} \sim \int_{1}^{x} \frac{\ln(t)}{t+1}$
as $\displaystyle x \rightarrow \infty$, where $\displaystyle f(x) \sim g(x)$ means $\displaystyle f(x)/g(x) \rightarrow 1$ as $\displaystyle x \rightarrow \infty$. I've thought about this for a while; is it just a standard kind of integration result or can you prove it?