lim(1/n的三次方+1）+（4/n的三次方+2）一直到+n的平方/n的三次方+n的极限

题目

lim(1/n的三次方+1）+（4/n的三次方+2）一直到+n的平方/n的三次方+n的极限

答案

设Sn = 1/(n³ + 1) + 4/(n³ + 2) + 9/(n³ + 3) + ...+ n²/(n³ + n) = Σ(k=1~n) k²/(n³ + k)
k²/(n³ + n) ≤ k²/(n³ + k) ≤ k²/n³
(1 + 4 + 9 + ...+ n²)/(n³ + n) ≤ Sn ≤ (1 + 4 + 9 + ...+ n²)/n³
lim(n-->∞) (1 + 4 + 9 + ...+ n²)/(n³ + n) = lim(n-->∞) (1/6)n(n+1)(2n+1)/(n³+n) = 1/3
lim(n-->∞) (1 + 4 + 9 + ...+ n²)/n³ = 1/6)n(n+1)(2n+1)/n³ = 1/3
∴lim(n-->∞) Sn = 1/3
==> lim(n-->∞) 1/(n³ + 1) + 4/(n³ + 2) + 9/(n³ + 3) + ...+ n²/(n³ + n) = 1/3