高中不等式证明题

来源：百度知道 编辑：UC知道 时间：2024/05/20 07:34:30
已知n为正整数，
求证：（1/2）*（3/4）*（5/6）*……[（2n-1)/2n]<1/[√（2n+1)]
（√-根号）
最好用放缩证明。数学归纳也可以

(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n] <
(2/3) * (4/5) * (6/7) * ...... [2n/(2n+1)]
不等式两边同乘以(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n]得：
{(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n]}^2 <
{(2/3) * (4/5) * (6/7) * ...... [2n/(2n+1)]} * {(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n]}
= (1/2) * (2/3) * (3/4) * (4/5) * (5/6) * ... * [(2n-1)/2n] * [2n/(2n+1)]
= 1/(2n+1)
{(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n]} > 0
因此{(1/2) * (3/4) * (5/6) * ...... [(2n-1)/2n]} < 1/[√（2n+1)]