灵鹊做题网一道数列放缩证明题,望大家不吝赐教!
来源：学生作业帮 编辑：灵鹊做题网作业帮 分类：数学作业 时间：2024/05/15 10:52:14
一道数列放缩证明题,望大家不吝赐教!
bn=1/(4n^2) .求证：当n≥1时,b1+(√2）b2+(√3)b3+.+(√n)bn
bn=1/(4n^2) .求证：当n≥1时,b1+(√2）b2+(√3)b3+.+(√n)bn
(√n)bn = (√n)/(4n^2) = (1/4) * n^(-3/2)
b1+(√2）b2+(√3)b3+.+(√n)bn
= (1/4) * (1^(-2/3) + 2^(-2/3) + 3^(-2/3) + ...+ n^(-2/3))
估算 n^(-2/3) 的上界
因为 1/√k - 1/√(k+1)
= (√(k+1)-√k) / (√k√(k+1))
= 1 / (√k√(k+1)(√(k+1)+√k))
> 1 / (√(k+1)√(k+1)(√(k+1)+√(k+1)))
= (1/2) * (k+1)^(-2/3)
所以 (k+1)^(-2/3) < 2 * (1/√k - 1/√(k+1))
b1+(√2）b2+(√3)b3+.+(√n)bn
= (1/4) * (1^(-2/3) + 2^(-2/3) + 3^(-2/3) + ...+ n^(-2/3))
< (1/4) * (1 + 2*(1/√1-1/√2) + 2*(1/√2-1/√3) + ...+ 2*(1/√(n-1)-1/√n))
= (1/4) * (1 + 2 - 2/√n)
< 3/4
b1+(√2）b2+(√3)b3+.+(√n)bn
= (1/4) * (1^(-2/3) + 2^(-2/3) + 3^(-2/3) + ...+ n^(-2/3))
估算 n^(-2/3) 的上界
因为 1/√k - 1/√(k+1)
= (√(k+1)-√k) / (√k√(k+1))
= 1 / (√k√(k+1)(√(k+1)+√k))
> 1 / (√(k+1)√(k+1)(√(k+1)+√(k+1)))
= (1/2) * (k+1)^(-2/3)
所以 (k+1)^(-2/3) < 2 * (1/√k - 1/√(k+1))
b1+(√2）b2+(√3)b3+.+(√n)bn
= (1/4) * (1^(-2/3) + 2^(-2/3) + 3^(-2/3) + ...+ n^(-2/3))
< (1/4) * (1 + 2*(1/√1-1/√2) + 2*(1/√2-1/√3) + ...+ 2*(1/√(n-1)-1/√n))
= (1/4) * (1 + 2 - 2/√n)
< 3/4