If `N = 1.2.3 .......m` (m is a fixed positive integer >2), then `1/log_2 N+1/log_3 N+..........1/log_m N` is equal to -

(1 + log_(n)m) * log_(mn) x is equal to

Let n=2006!. Then 1/(log_(2)n)+1/(log_(3)n)+…+1/(log_(2006)n) is equal to :

If ngt1 then prove that 1/(log_2 n )+1/(log_3 n)+.................+1/log_(53) n =1/(log_(53!) n

Express M in terms of N: 1/2 log_(3) M +log_(3)N=1

Let x in(1,oo) and n be a positive integer greater than 1 . If f_n (x) =n/(1/log_2x+1/log_3x+...+1/log_nx) , then (n!)^(f_n (x)) equals to