Answer
$-\dfrac{1}{\ln 2}$
Work Step by Step
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, $=-\dfrac{1}{\ln 2}$
