Answer
Diverges
Work Step by Step
$$\sum ^{\infty }_{n=1}\ln \left( \dfrac {n+1}{n}\right) =\sum ^{\infty }_{n=1}\left( \ln \left( n+1\right) -\ln n\right) =\left( \ln 2-\ln 1\right) +\left( \ln 3-\ln 2\right) ...\ln \left( n+1\right) -\ln n==\ln \left( n+1\right) -\ln 1=\ln \left( n+1\right) =\infty $$
$$\Rightarrow \sum ^{\infty }_{n=1}\ln \left( \dfrac {n+1}{n}\right) $$ diverges