If `n=|__2002,` evaluate `1/(log_2n)+1/(log_3n)+1/(log_4n)+..........+1/(log_2002 n).`

If n=|__(2020) , then 1/(log_(2)n)+1/(log_(3)n)+1/(log_(4)n)+………..+1/(log_(2020)n)=

If n=(2017)! , then what is (1)/(log_(2)n)+(1)/(log_(3)n)+(1)/(log_(4)n)+....+(1)/(log_(2017)n) equal to?

Show that: 1/(log_2n)+1/(log_3n)+1/(log_4n)+…+1/(log_43n)=1/(log_(43!)n)

(1)/(log_(2)(n))+(1)/(log_(3)(n))+(1)/(log_(4)(n))+....+(1)/(log_(43)(n))