Question

各项均为正数且公差为1的等差数列an,前n项和为Sn,求limSn/an×an+1

Answer 1

设首项为a1,d=1

前n项和为Sn=na1+n(n-1)d/2=na1+n(n-1)/2

an=a1+(n-1)d=a1+n-1

limSn/an×an+1=lim[na1+n(n-1)/2]/(a1+n-1)(a1+n-1)+1

=lim[a1/n+1/2-1/(2n)]/[(a1/n+1-1/n)^2+1/(n^2)]

=1/2