等差数列{an}前n项和为sn,则lim（n趋向无穷）sn/（an^2）=?

Question

等差数列{an} 前n项和为sn,则lim（n趋向无穷）【sn/（an^2）】=?

Answer 1

a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,
s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),
当d=0时,a(n)=a,s(n)=na.
当d=0,a=0时,s(n)/[a(n)]^2没有意义.
当d=0,a不等于0时,s(n)/[a(n)]^2=n/a,lim(n->无穷大){s(n)/[a(n)]^2}=(n->无穷大){n/a}=正无穷大.
d不等于0时,
s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,
lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷大）{[d/2+(a-d/2)/n]/[d+(a-d)/n]^2}=(d/2)/d^2=1/(2d)