h***n
发帖数: 276 | 1 已知 {ai},{bi},{xi}和{yi}都是正的数列，其级数收敛
另外 sum{i=1,...,infinity}ai*xi=sum{i=1,...,infinity}bi*yi
还有对于任意n,m>=1,都有 an<=bm
问题是比较 sum{i=1,...,infinity}xi和sum{i=1,...,infinity}yi.前者一定大于后者
吗？
多谢！ | B****n
发帖数: 11290 | 2 if sup{ai}>0 and inf{bi}>0, then the inequality holds.
【在 h***n 的大作中提到】
: 已知 {ai},{bi},{xi}和{yi}都是正的数列，其级数收敛
: 另外 sum{i=1,...,infinity}ai*xi=sum{i=1,...,infinity}bi*yi
: 还有对于任意n,m>=1,都有 an<=bm
: 问题是比较 sum{i=1,...,infinity}xi和sum{i=1,...,infinity}yi.前者一定大于后者
: 吗？
: 多谢！
| e**********n
发帖数: 359 | 3 I guess sum bi and sum ai converge, which are given conditions. So inf{bi} =
inf{ai} = 0.
Under these conditions, this inequality is not true. Counter example:
a1 = 1, b1 = 11/10, x1 = 1, y1 = 8/5
a2 = 76/45, b2 = 11*a2/10, x2 = 1, y2 = 1/2
and the remaining numbers can be made arbitrarily small, and let their sums
not only converge but also
be arbitrarily small.
Obviously y1+y2>x1+x2, and all the conditions are satisfied.
【在 B****n 的大作中提到】
: if sup{ai}>0 and inf{bi}>0, then the inequality holds.
| t*******r
发帖数: 6 | 4 原来的条件会不会是 a_n<=b_m, 对所有的 n>=m ?
=
sums
【在 e**********n 的大作中提到】
: I guess sum bi and sum ai converge, which are given conditions. So inf{bi} =
: inf{ai} = 0.
: Under these conditions, this inequality is not true. Counter example:
: a1 = 1, b1 = 11/10, x1 = 1, y1 = 8/5
: a2 = 76/45, b2 = 11*a2/10, x2 = 1, y2 = 1/2
: and the remaining numbers can be made arbitrarily small, and let their sums
: not only converge but also
: be arbitrarily small.
: Obviously y1+y2>x1+x2, and all the conditions are satisfied.
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