已知数列an是等差数列,且a1不等于0,Sn为这个数列的前n项和,求limnan/Sn.limSn+Sn-1/Sn+Sn-1

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-1趋于无穷,所以1/(n-1)趋于0所以极限=2d/d=22、原式=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=...