University Calculus: Early Transcendentals (3rd Edition)

Here, $s_n=(\ln \sqrt 2 -\ln \sqrt 1)+(\ln \sqrt 3 -\ln \sqrt 2)+(\ln \sqrt 4 -\ln \sqrt 3)+.....(\ln \sqrt {n+1} -\ln \sqrt n)=\ln \sqrt {n+1}$ Thus, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} \ln (\sqrt {n+1})=\infty$ Now, $s_n \to \infty$ as $n \to \infty$, so, the given series diverges.