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

}

" />