Question

what is $\frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+\frac{1}{{\log }_{4}N}\dots +\frac{1}{{\log }_{100}N}$ equal to $\left(N\ne 1\right)$ ?

Answer 1

$A=\frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+\frac{1}{{\log }_{4}N}\dots +\frac{1}{{\log }_{100}N}$

$=\frac{\log 2}{\log N}+\frac{\log 3}{\log N}+\frac{\log 4}{\log N}\dots +\frac{\log 100}{\log N}$

$=\frac{\log 2+\log 3+\log 4+......+\log N}{\log N}$

$=\frac{\log (2\times 3\times \mathrm{4....}\times 100)}{\log N}$

$=\frac{\log (100!-1)}{\log N}$

$={\log }_N(100!-1)$