Question

{an}是等差数列,a1不等于0,Sn是它前n项的和.

Answer 1

(Sn+S(n+1))/(Sn+S(n-1))

=[(na1+n(n-1)d/2)+((n+1)a1+n(n+1)d/2)]/[(na1+n(n-1)d/2)+((n-1)a1+(n-1)(n-2)d/2)]

=((2n+1)a1+n^2d)/((2n-1)a1+(n-1)^2d)

所以lim(n趋于无穷)=1