发布网友 发布时间:2024-10-22 01:14
共1个回答
热心网友 时间:2024-11-09 10:08
x(n+1)=[(2n-1)/(2n+2)]x(n),
(2n+2)x(n+1)=(2n-1)x(n),
(2n+2)(2n+1)(2n)x(n+1)=(2n+1)(2n)(2n-1)x(n),
{(2n+1)(2n)(2n-1)x(n)}是首项为(2+1)(2)(1)x(1)=3的常数数列.
(2n+1)(2n)(2n-1)x(n)=3,
x(n) = 3/[(2n+1)(2n)(2n-1)]
=(3/2){1/[(2n+1)(2n-1)n]}
<=(3/2){ 1/[(2n+1)(2n-1)] }
=(3/4)[ 1/(2n-1) - 1/(2n+1)]
s(n)=x(1)+x(2)+...+x(n-1)+x(n)
<=(3/4)[ 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/(2n-3) - 1/(2n-1) + 1/(2n-1) - 1/(2n+1)]
=(3/4)[1 - 1/(2n+1)]
<3/4
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