When n tends to infinity, how to find the limit of 1 / N ^ 3 + 2 ^ 2 / N ^ 3 +... + (n-1) ^ 2 / N ^ 3

There is a summation formula
1+2^2+.+n^2=n*（n+1）*(2n+1)/6;
So after the same score, we can get:
(n-1)*n*(2n-1)/（6n^3)
The limit is one third

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