## Thomas' Calculus 13th Edition

$-\dfrac{1}{\ln 2}$
The nth partial sums are: $s_n=(\dfrac{1}{\ln 3}-\dfrac{1}{\ln 2})+(\dfrac{1}{\ln 4}-\dfrac{1}{\ln 3})+.....+(\dfrac{1}{\ln (n+2)}-\dfrac{1}{\ln (n+1)})=[-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})]$ Now, $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})]=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2}] +\lim\limits_{n \to \infty} \dfrac{1}{ \ln (n+2)})]$ or, $=-(\dfrac{1}{\ln 2})+0$ or, $=-\dfrac{1}{\ln 2}$