已知等差数列{an}的首项a1=1,公差d=1,前n项和为Sn,bn=1/Sn （1）求bn（2）求证：b1+b2+……+bn

(1) Sn=[2a1+(n-1)d]*n/2=[2*1+n-1]*n/2=n(n+1)/2所以bn=1/Sn=2/n(n+1)(2) bn=2[1/n-1/(n+1)]b1+b2+...+bn=2(1-1/2)+2(1/2-1/3)+...+2[1/n-1/(n+1)=2[1-1/(n+1)]=2-2/(n+1)