Let $n \in \mathbb { N } , n \geqslant 2$. Justify that $$\pi ( \sqrt { n } ) \leqslant \sqrt { n } < \frac { n } { \ln ( n ) }$$ then deduce that $$\pi ( n ) \leqslant 4 \frac { \ln ( n ) } { n }$$ One may note that $2 > \ln ( 4 )$.
Let $n \in \mathbb { N } , n \geqslant 2$. Justify that
$$\pi ( \sqrt { n } ) \leqslant \sqrt { n } < \frac { n } { \ln ( n ) }$$
then deduce that
$$\pi ( n ) \leqslant 4 \frac { \ln ( n ) } { n }$$
One may note that $2 > \ln ( 4 )$.