Evaluate :
`lim_( n ->oo ) [( 1 + 2 + 3 --- ( n -1 ) )/n^2 ]`

Evaluate: lim_ (n rarr oo) (1 * 2 + 2 * 3 + 3 * 4 + ... + n (n + 1)) / (n ^ (3))

evaluate lim_ (n rarr oo) [(1) / (3) + (1) / (3 ^ (2)) + (1) / (3 ^ (2)) + ......... + (1) / (3 ^ (n))]

lim_ (n rarr oo) (1 + 2 + 3 * -n) / (n ^ (2))

Evaluate: lim_ (n rarr oo) (1 ^ (4) + 2 ^ (4) + 3 ^ (4) + ... + n ^ (4)) / (n ^ (5)) - lim_ (n rarr oo) (1 ^ (3) + 2 ^ (3) + ... + n ^ (3)) / (n ^ (5))

Evaluate: lim_ (n rarr oo) (1 ^ (4) + 2 ^ (4) + 3 ^ (4) + ... + n ^ (4)) / (n ^ (5)) - lim_ (n rarr oo) (1 ^ (3) + 2 ^ (3) + ... + n ^ (3)) / (n ^ (5))

The value of lim_(x to oo) (1 + 2 + 3 … + n)/(n^(2)) is

Evaluate lim_ (n rarr oo) [(1+ (1) / (n)) 91+ (2) / (n)) .... (1+ (n) / (n))] ^ ((1) /(a))

Using Sandwich theorem, evaluate lim_ (n rarr oo) ((1) / (1 + n ^ (2)) + (1) / (2 + n ^ (2)) + ...... + (n) / (n + n ^ (2)))

lim_ (n rarr oo) (1 + 2 + 3 + ...... + n) / (n ^ (2))