Question

已知数列an=1/(3^n-n-1)的前n项和为Sn,证明：Sn＜2对任意n∈N+都成立.

Answer 1

a1=1/(3-1-1)=1

a(n+1)/an=(3ⁿ-n-1)/[3^(n+1)-(n+1)-1]

=(1/3)[3^(n+1)-3n-3]/[3^(n+1)-(n+1)-1]

=(1/3)[3^(n+1)-(n+1)-1-2n-1]/[3^(n+1)-(n+1)-1]

=(1/3){1 -(2n+1)/[3^(n+1)-(n+1)-1]}

=1/3 - (2n+1)/[3^(n+2)-3(n+1)-3]

(2n+1)/[3^(n+2)-3(n+1)-3]>0

1/3 - (2n+1)/[3^(n+2)-3(n+1)-3]