Question

已知等比数列an的公比大于1,s3=39.且a1,2/3a2,1/3a3成等差,设bn=1/a1+1/a2+1/a3+...+1/an,求证bn<1/2

Answer 1

若an的公比=q
a1,2/3a2,1/3a3成等差, 有

2/3 *a2 *2 = a1 + 1/3 a3

即 4 a2 = 3 a1 + a3

4 a1 q = 3 a1 + a1 q²
所以 4q = 3 + q²

(q-1)(q-3)=0
因为q>1, 所以q-1>0

所以 q=3

a1(1+q+q²) = a1(1+3+9) = 39
所以 a1=3

bn = 1/a1 *(1+ 1/q+ 1/q² + ... + 1/q^(n-1))
= 1/3 * (1+ 1/3 + 1/9 + ... + 1/3^(n-1))
= 1/3 *(1-(1/3)^n)/(1-1/3)
= 1/2 - 1/2 *(1/3)^n
< 1/2