求證：
[數學]數學題..(62)
第1頁(共1頁)
由 galaxylee 於 星期一 十月 24, 2005 9:59 pm
原式=(1/2)-(1/3)+(1/4)-(1/5)+...+(1/2004)-(1/2005)
令p=(1/2)-(1/3)+(1/4)-(1/5)+...+(1/2004)-(1/2005)
1-p
=1-(1/2)+(1/3)-(1/4)+(1/5)+...-(1/2004)+(1/2005)
=1+(1/2)+(1/3)+...+(1/2005)-2[(1/2)+(1/4)+...+(1/2004)]
=1+(1/2)+(1/3)+...+(1/2005)-[1+(1/2)+...+(1/1002)]
=(1/1003)+(1/1004)+...+(1/2005)
>(1/2005)*1003
=(1003/2005)
所以p<1-(1003/2005)=(1002/2005)
令p=(1/2)-(1/3)+(1/4)-(1/5)+...+(1/2004)-(1/2005)
1-p
=1-(1/2)+(1/3)-(1/4)+(1/5)+...-(1/2004)+(1/2005)
=1+(1/2)+(1/3)+...+(1/2005)-2[(1/2)+(1/4)+...+(1/2004)]
=1+(1/2)+(1/3)+...+(1/2005)-[1+(1/2)+...+(1/1002)]
=(1/1003)+(1/1004)+...+(1/2005)
>(1/2005)*1003
=(1003/2005)
所以p<1-(1003/2005)=(1002/2005)
-
galaxylee - 副教授
- 文章: 555
- 註冊時間: 2005-07-18
第1頁(共1頁)