Question

Any Math Aryabhatta .....plzz Solve it ​

Answer 1

Explanation:

lim n->infi. 1/1+n² + 2/2+n² + 3/3+n² + ... + n/n+n²

=> just divide up and down by n²

=> lim n-> inf. 1/n²/(1/n² + 1) + (2/n²/(2/n²+1)) + .... + (1/n/(1/n + 1))

=> put n value.

=> 1/infin = 0

=> 2/inf = 0

.

.

=> n/inf = 0

=> 0/(0+1) + 0/(0+1) + 0/(0+1) + ... + 0/(0+1)

=> 0/1 + 0/1 + 0/1 +... + 0/1

=> 0