Question

一道数列问题，第21题

Answer 1

a(1)=-1/2,
2a(n+1) = a(n) - 1,
2a(n+1) + 2 = a(n) + 1,
a(n+1)+1 = (1/2)[a(n)+1],
{a(n)+1}是首项为a(1)+1=1/2,公比为1/2的等比数列。
a(n)+1 = (1/2)(1/2)^(n-1) = 1/2^n.

a(n) = (1/2)^n - 1.

b(n) = 1/[2^na(n)a(n+1)] = 1/[(1-2^n)(1/2^(n+1) - 1)] = 1/[(2^n - 1)(1 - 1/2^(n+1))]
= 2^(n+1)/[(2^n-1)(2^(n+1)-1)]
= 2[1/(2^n-1) - 1/(2^(n+1)-1)],

s(n) = b(1) + b(2) + ... + b(n-1) + b(n)
= 2[1/(2-1) - 1/(2^2-1) + 1/(2^2-1) - 1/(2^3-1) + ... + 1/[2^(n-1)-1] - 1/[2^n-1] + 1/[2^n-1] - 1/[2^(n+1)-1] ]
= 2{1/(2-1) - 1/[2^(n+1)-1] }
= 2 - 2/[2^(n+1)-1]
< 2