已知数列{an}中,a1=1,前n项和为Sn,且an-a(n+1)+1=0.计算lim(1/S1+1/S2+...+1/Sn).

Question

已知数列{an}中,a1=1,前n项和为Sn,且an-a(n+1)+1=0.计算lim(1/S1+1/S2+...+1/Sn).

Answer 1

根据题意,{an}为等差数列,且d=1,所以an=n,Sn=[n(n+1)]/2
1/Sn=2(1/n-1/(n+1))(裂项相消法)
1/S1+1/S2+...+1/Sn=2[1-1/(n+1)]
lim(1/S1+1/S2+...+1/Sn)=2