Question

无穷等比数列﹛an﹜的前n项和Sn，首项a1，公比q满足|q|＜1，所有项的和为S

求lim﹙S1＋S2＋…＋Sn－n S﹚

Answer 1

所有项的和S是a1/(1-q)
前n项之和是Sn是a1(1-q^n)/(1-q)
Sn-S=-a1(q^n)/(1-q)
现在就是求 ﹙S1＋S2＋…＋Sn－n S﹚
Sn-S就是以-a1q/(1-q)为首项，q为公比的等比数列
lim﹙S1＋S2＋…＋Sn－n S﹚=-a1q/(1-q)^2

Answer 2

Sn=a1(qⁿ-1)/(q-1)

S1+S2+...+Sn
=[a1/(q-1)](q+q²+...+qⁿ-n)
=[a1/(q-1)][q(qⁿ-1)/(q-1) -n]
=a1q(qⁿ-1)/(q-1)² -na1/(q-1)
n->+∞，qⁿ->0 S1+S2+...+Sn->-a1q/(q-1)²-na1/(q-1)

Sn=a1(qⁿ-1)/(q-1) n->+∞，Sn->-a1/(q-1)
nS=-na1/(q-1)

lim(S1+S2+...+Sn-nS)
=-a1q/(q-1)²-na1/(q-1)+na1/(q-1)
=-a1q/(q-1)²