Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is D $1$

Given that $N=n!(n\in N,n>2)$
${\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....{\left({\log }_{n}N\right)}^{-1}={\log }_{N}2+{\log }_{N}3+....+{\log }_{N}n={\log }_{N}2.3\cdot \cdot \cdot \cdot n$
$\Rightarrow {\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....{\left({\log }_{n}N\right)}^{-1}={\log }_{N}2.3\cdot \cdot \cdot \cdot n$
$\Rightarrow {\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....{\left({\log }_{n}N\right)}^{-1}={\log }_{N}n!={\log }_{N}N=1$
${lim}_{N\to \infty }[{\left({\log }_{2}N\right)}^{-1}+{\left({\log }_{3}N\right)}^{-1}+.....{\left({\log }_{n}N\right)}^{-1}]={lim}_{N\to \infty }1=1$
Ans: D