Limit as n approaches infinity of 1/n[ (1/n) ^9 + (2/n) ^9 + (3/n) ^9 + ... + (n/n) ^9

As each term is convergent thus sum is also convergent ... Therefore as n tends to infinity, the each term converges to 0. thus sum is also converges to 0+0+0+0 + ... n times=0