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

Prove that: 1/(log_2N)+1/(log_3N)+1/(log_4N)+…+1/(log_(2011)N)=1/(log_(2011!)N)

If n >1,t h e np rov et h a t 1/((log)_2n)+1/((log)_3n)++1/((log)_(53)n)=1/((log)_(53 !)n)dot

(1)/("log"_(2)n) + (1)/("log"_(3)n) + (1)/("log"_(4)n) + … + (1)/("log"_(43)n)=

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

The sum of the series: 1/((log)_2 4)+1/((log)_4 4)+1/((log)_8 4)++1/((log)_(2n)4) is (n(n+1))/2 (b) (n(n+1)(2n+1))/(12) (c) (n(n+1))/4 (d) none of these

Prove that ((log)_a N)/((log)_(a b)N)=1+(log)_a b

Prove the following identities: (a) (log_(a) n)/(log_(ab) n) = 1+ log_(a) b" "(b) log_(ab) x = (log_(a) x log_(b) x)/(log_(a) x + log_(b) x) .

1+(log_e n)^2 /(2!) + (log_e n )^4 / (4!)+...=

In a binomial distribution B(n , p=1/4) , if the probability of at least one success is greater than or equal to 9/(10) , then n is greater than (1) 1/((log)_(10)^4-(log)_(10)^3) (2) 1/((log)_(10)^4+(log)_(10)^3) (3) 9/((log)_(10)^4-(log)_(10)^3) (4) 4/((log)_(10)^4-(log)_(10)^3)

In a binomial distribution B(n , p=1/4) , if the probability of at least one success is greater than or equal to 9/(10) , then n is greater than (1) 1/((log)_(10)^4-(log)_(10)^3) (2) 1/((log)_(10)^4+(log)_(10)^3) (3) 9/((log)_(10)^4-(log)_(10)^3) (4) 4/((log)_(10)^4-(log)_(10)^3)