Let $n \in \mathbb { N } ^ { * }$. Verify that $$\frac { 2 n \ln ( 2 ) } { \ln ( 2 n ) } - 1 \geqslant \frac { n \ln ( 2 ) } { \ln ( 2 n ) }$$ then deduce that $$\pi ( 2 n ) \geqslant n \frac { \ln ( 2 ) } { \ln ( 2 n ) }$$