一个极限证明
求证$\lim_{n->infty}1/{2^n\sqrt(n)}\sum_{j=1}^n({:(n),(j):})\sqrt(j)=1/{\sqrt(2)}$
来自mymathforum. 补充说明
$({:(n),(j):})$我们通常写成$C_n^j$,也就是组合数 mathe怎么证明的？ 这个应该不是很难的 肚老大
最近很忙啊
在论坛发言少了
呵呵 **** Hidden Message ***** So we may use squeeze thm by finding the lower and upper bound of the limit as 1/square root 2.
2^n can be written as sum of (n,j) from 0 to inf. we may assume it is from 1 to inf cause the difference is tiny when n goes inf.
also since square root of j is concave down, we can apply the jesens inequality. then the original limit will be small or equal to sqr ( sum (n,j)*j/ sum (n,j)) * 1/ sqr(n)
sum(n,j) *j is n*2^(n-1) simplifed to 1/ sqr(2)
To find the lower bound, i am trying to adjust sqr(j) to 1/sqr(j) in order to apply the inequality again with concave up function. but there is some "re-index" problem need further clearifying. But i guess it's almost there, well, hopefully....
回复 1# mathe 的帖子
我有一个证法： 放缩法左边式子的$=1/2(sqrt(j)+sqrt(n-j))$,而
$sqrt(n/2)<=1/2(sqrt(j)+sqrt(n-j))<=1/2sqrt(2n)$
分别代进去，都得$sqrt(2)/2$ 左边弄错了:L 学习ing
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