等差数列列｛an｝的前n项和为Sn,公差为2,则lim(n→∞)Sn/an^2=?

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等差数列列｛an｝的前n项和为Sn,公差为2,则lim(n→∞)Sn/an^2=?
等差数列列｛an｝的前n项和为Sn,公差为2,则lim(n→∞)Sn/an^2=?

等差数列列｛an｝的前n项和为Sn,公差为2,则lim(n→∞)Sn/an^2=?
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),
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)
d=2
lim(n->无穷大){s(n)/[a(n)]^2}=0.25

Sn是n的一次式，an^2是n的二次式，极限是0

lim(n→∞)[n*a1^2+n*(n-1)*d]/2*[a1+(n-1)d]^2=1/d=1/2

4分之1

张家琛wd 正确

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