求极值 lim(n->无限) [n(n+1)(2n+1)]/6n^2

lim(n→∝) [n(n+1)(2n+1)]/6n^2
=lim(n→∝) [(n+1)(2n+1)]/6n
=lim(n→∝) (2n^2+3n+1)/6n
=lim(n→∝) (n/3+1/2+1/6n)
lim(n→∝) (1/2+1/6n)=1/2
lim(n→∝) (n/3)=∝
所以lim(n→∝) [n(n+1)(2n+1)]/6n^2=∝

lim[n(n+1)(2n+1)]/6n^2
分子分母同时除以n^3
=lim[(1+1/n)(2+1/n)]/(6/n)
其中分母[(1+1/n)(2+1/n)]趋于2，分子6/n趋于0
所以整体趋于∞

也即lim[n(n+1)(2n+1)]/6n^2
=lim[(1+1/n)(2+1/n)]/(6/n)=∞

分子比分母高了一次，所以极限是无穷大。另外你注意到此题的特殊性了吗：[n(n+1)(2n+1)]/6=1^2+2^2+3^2+…+n^2.