Question

求数列{n²/2ⁿ}的极限

Answer 1

极限为0
2^n=(1+1)^n,由二项式定理得2^n=1+n+n(n-1)/2+n(n-1)(n-2)/6+...
因为展开式内每一项都为正,故2^n>1+n+n(n-1)/2+n(n-1)(n-2)/6
∴0<n²/2^n<n²/[1+n+n(n-1)/2+n(n-1)(n-2)/6]
∵n²/[1+n+n(n-1)/2+n(n-1)(n-2)/6]
=1/(An+B+C/n+D/n²),其中A,B,C,D是分母各项的系数且A≠0
当n→∞时,分母极限为∞,因此lim(n→
∞)n²/[1+n+n(n-1)/2+n(n-1)(n-2)/6]=0
夹逼定理得lim(n→∞)n²/2^n=0

Answer 2

n²/2ⁿ=e∧（2logn-nlog2）=e∧（-∞）=0