Answer
$-\dfrac{1}{\ln 2}$
Work Step by Step
Here, we have:
$ 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)})$
Thus, we have the sum:
$\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})]=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2}] +\lim\limits_{n \to \infty} \dfrac{1}{ \ln (n+2)})]=-\dfrac{1}{\ln 2}+0=-\dfrac{1}{\ln 2}$
