Question

等比数列{an}的首项是a1,公比是q,前n项之和为Sn,所有项的和为S,则lim(S1+S2+...+Sn-nS)=_

Answer 1

S1=a1(1-q)/(1-q),S2=a1(1-q^2)/(1-q),...,Sn=a1(1-q^n)/(1-q).

S1+S2+...+Sn=[a1/(1-q)]*[1-q+1-q^2+...+1-q^n]

=[a1/(1-q)]*[n-q(1-q^n)/(1-q)]

=na1/(1-q)-a1q(1-q^n)/(1-q)^2,

nS=na1/(1-q),

(S1+S2+...+Sn)-nS=-a1q(1-q^n)/(1-q)^2,

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