Answer
$\dfrac{1}{\ln (2)}$
Work Step by Step
Here, we have
$S_n= \Sigma_{k=2}^{n} \dfrac{\ln ((k+1)k^{-1})}{\ln k \ln (k+1)}$
or, $= \dfrac{\ln(3/2)}{\ln (2) \ln (3)}+\dfrac{\ln(4/3)}{\ln (4) \ln (3)}+...+\dfrac{\ln [(n+1)/n]}{\ln (n) \ln (n+1)}$
or, $= \dfrac{\ln 3-\ln 2}{\ln (2) \ln (3)}+\dfrac{\ln 4-\ln 3}{\ln (4) \ln (3)}+...+\dfrac{\ln (n+1)-\ln n}{\ln (n) \ln (n+1)}$
or, $= \dfrac{1}{\ln 2}-\dfrac{1}{\ln (n+1)}$
Now, $S=\lim\limits_{n \to \infty} \dfrac{1}{\ln 2}-\lim\limits_{n \to \infty}\dfrac{1}{\ln (1+1/n)} $
or, $S= \dfrac{1}{\ln {2}}-0$
or, $= \dfrac{1}{\ln (2)}$
