一个极限证明

mathe

求证 $\lim_{n->infty}1/{2^n\sqrt(n)}\sum_{j=1}^n({:(n),(j):})\sqrt(j)=1/{\sqrt(2)}$ 来自mymathforum.

mathe

补充说明 $({:(n),(j):})$我们通常写成$C_n^j$,也就是组合数

northwolves

mathe怎么证明的？

mathe

这个应该不是很难的

无心人

肚老大 最近很忙啊 在论坛发言少了 呵呵

mathe

游客，如果您要查看本帖隐藏内容请回复

shangwen_ren

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....

wayne

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$sqrt(2)/2$

wayne

左边弄错了

jackie

学习ing