数列放缩已知an=n^2,求证1/a1+1/a2+…+1/an
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 14:45:51
数列放缩已知an=n^2,求证1/a1+1/a2+…+1/an
数列放缩
已知an=n^2,求证1/a1+1/a2+…+1/an
数列放缩已知an=n^2,求证1/a1+1/a2+…+1/an
法2:
当n=1或2时,成立;
n>2时,
1/a1+1/a2+...+1/an
=1+1/4+1/3^2+...+1/n^2
<5/4+1/[(3-1)*3]+...+1/[(n-1)*n]
=5/4+[1/2-1/3]+...+[1/(n-1)-1/n]
=7/4-1/n
<7/4
原式=1/1+1/4+1/9+1/16+。。。。。。。+1/N^2
原式<1+1/(1*3)+1/(2*4)+1/(3*5).........+1/[(n-1)(n+1)]
=1+1/2[1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+.......+1/(n-1)-1/(n+1) ]
...
全部展开
原式=1/1+1/4+1/9+1/16+。。。。。。。+1/N^2
原式<1+1/(1*3)+1/(2*4)+1/(3*5).........+1/[(n-1)(n+1)]
=1+1/2[1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+.......+1/(n-1)-1/(n+1) ]
=1+1/2{1+1/2-[1/(n+1)+1/n]}
=1+1/2*3/2-1/2[1/(n+1)+1/n]
=7/4-1/2[1/(n+1)+1/n]
<7/4
收起