证明 0.0999<1/10^2+1/11^2+...+1/1000^2<0.111
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解决时间 2021-02-04 02:55
- 提问者网友：黑米和小志
- 2021-02-03 10:20
证明 0.0999<1/10^2+1/11^2+...+1/1000^2<0.111
最佳答案
- 五星知识达人网友：北城痞子
- 2021-02-03 11:53
[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]>[1/(10x11)+1/(11x12)+1/(12x13)+.....+1(1000x1001)]
1/(10x11)+1/(11x12)+1/(12x13)+.....+1(1000x1001)
=(1/10)-(1/11)+(1/11)-(1/12)+.......+(1/1000)-(1/1001)
=(1/10)-(1/1001)=0.099
[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]<[1/(9x10)+1/(10x11)+1/(11x12)+.....+1(999x1000)]
=(1/9)-(1/10)+(1/10)-(1/11)+.......+(1/999)-(1/1000)
=(1/9)-(1/1000)
=991/9000=0.110
0.099＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.110
1/(10x11)+1/(11x12)+1/(12x13)+.....+1(1000x1001)
=(1/10)-(1/11)+(1/11)-(1/12)+.......+(1/1000)-(1/1001)
=(1/10)-(1/1001)=0.099
[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]<[1/(9x10)+1/(10x11)+1/(11x12)+.....+1(999x1000)]
=(1/9)-(1/10)+(1/10)-(1/11)+.......+(1/999)-(1/1000)
=(1/9)-(1/1000)
=991/9000=0.110
0.099＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.110
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