evaluate lim x ->infinite 1.2+2.3+3.4+..+n(n+1)/n3

Question

i couldnt get the answer cause i dont know the value of sigma n(n+1)

Answer 1

(n(n + 1))/n3

 

can be rewritten as (n2 + n)/n3.

 

Divide the numerator and denominator by n3:

 

(1/n + 1/n2)/1

 

For large values of n, the expression becomes:

 

(0 + 0)/1 or 0

Dividing 1 by large numbers yields small quotients whose values approach 0.