m*********s
发帖数: 5 | 1 Hi all,
Given k_n*|x_n| -> 0 and k_n -> 0 as n -> infinity. Prove that
k(n)*sup_{i<=n}|x_i| -> 0 as n -> infinity.
What I have so far:
If {x_n} is bounded, it's easy.
If {x_n} is not bounded, then for all positive n, there exists N >= n such
that
sup_{i<=N}|x_i| = |x_N| (take N to be the smallest integer >= n such that
sup_{i<=n}|x_i| <= |x_N|; thus j |x_j| < sup_i{i<=n}|x_i| <= |x_N|.)
Then for all episilon > 0, there exists a positive integer N such that
k(N)*sup_{i<=N}|x_i| = k(N)*|x_ | H****h
发帖数: 1037 | 2 不一定成立。考虑这样地情况：偶数的k_n都等于0，而偶数的x_n则非常地大。
【在 m*********s 的大作中提到】
: Hi all,
: Given k_n*|x_n| -> 0 and k_n -> 0 as n -> infinity. Prove that
: k(n)*sup_{i<=n}|x_i| -> 0 as n -> infinity.
: What I have so far:
: If {x_n} is bounded, it's easy.
: If {x_n} is not bounded, then for all positive n, there exists N >= n such
: that
: sup_{i<=N}|x_i| = |x_N| (take N to be the smallest integer >= n such that
: sup_{i<=n}|x_i| <= |x_N|; thus j |x_j| < sup_i{i<=n}|x_i| <= |x_N|.)
: Then for all episilon > 0, there exists a positive integer N such that
| m*********s
发帖数: 5 | 3 谢谢Health. I guess I can only show liminf_{n->infin} k_n*sup_{i<=n}|x_i| =
0. Here is a counter-example: k_n = 1/n if n odd, 0 if n even; x_n = 0 if n
odd, n+1 if n even. Then k_n*x_n = 0, and k_n -> 0 as n -> infinity, but
k_n*sup_{i<=n}|x_i| = 0 if n even, 1 if n odd.
【在 H****h 的大作中提到】
: 不一定成立。考虑这样地情况：偶数的k_n都等于0，而偶数的x_n则非常地大。
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