Question

$\mathrm{If}N=n!(n\in N,n>2)\mathrm{then}\left({\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....+{\left({\log }_{n}N\right)}^{-1}\right]\mathrm{is}$

Answer 1

${\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....+{\left({\log }_{n}N\right)}^{-1}={\log }_{N}2+{\log }_{N}3+.......+{\log }_{N}n={\log }_N\left(2.3.4......n\right)={\log }_N\left(n!\right)={\log }_{N}N=1$