Prove that
As x goes to infinity the sum
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
}
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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
as , where means as . I've thought about this for a while; is it just a standard kind of integration result or can you prove it?