1/a(n+1) = f[a(n)] = {4[a(n)]^2 + 1}^(1/2)/a(n), a(n) > 0.
1/[a(n+1)]^2 = {4[a(n)]^2 + 1}/[a(n)]^2 = 1/[a(n)]^2 + 4,
{1/[a(n)]^2 }是首项为1/[a(1)]^2 = 1, 公差为4的等差数列.
1/[a(n)]^2 = 1 + 4(n-1) = 4n-3
1/a(n) = (4n-3)^(1/2),
a(n) = (4n-3)^(-1/2).
[a(n)]^2 = 1/(4n-3).
s(n) = [a(1)]^2 + [a(2)]^2 + ... + [a(n]^2 = 1 + 1/5 + ... + 1/(4n-3),
s(2n+1) = 1 + 1/5 + ... + 1/(4n-3) + 1/[4(n+1) -3] + 1/[4(n+2) - 3] + ... + 1/[4(2n+1)-3],
b(n) = 1/(4n+1) + 1/(4n+5) + ... + 1/(8n+1)
= 1/(4n+1) + 1/[4(n+1)+1] + ... + 1/[4(n+n) + 1],
b(n+1) = 1/[4(n+1) + 1] + 1/[4(n+2) + 1] + ... + 1/[4(n+1+n-1) + 1] + 1/[4(n+1+n) + 1] + 1/[4(n+1+n+1) + 1]
= 1/[4(n+1) + 1] + ... + 1/[4(n+n) + 1] + 1/[4(n+1+n) + 1] + 1/[4(n+1+n+1) + 1],
b(n+1) - b(n) = 1/[4(n+1+n) + 1] + 1/[4(n+1+n+1) + 1] - 1/(4n+1)
= 1/(8n+5) + 1/(8n+9) - 1/(4n+1)
< 1/(8n+2) + 1/(8n+2) - 1/(4n+1)
= 2/(8n+2) - 1/(4n+1)
= 1/(4n+1) - 1/(4n+1)
= 0.
{b(n)}确实是单调递减滴.
