作者znmkhxrw (QQ)
看板Math
标题Re: [微积] 证明数列递增且有界
时间Mon May 30 19:41:46 2011
※ 引述《KOREALee (韩国最高)》之铭言:
: 标题: [微积] 证明数列递增且有界
: 时间: Mon May 30 19:13:49 2011
:
: Let the sequence
:
:
: 1 n
: An = (1+ ---)
: n
:
:
: 1.show that An is increasing
:
: 2.show that An is bounded
1
1. By comparison with ── , we find when N >= No
n^2
∞ 1
Σ ─── is conv.
n=N n!
Since
∞ 1
Σ ─── is conv. (by p-series)
n=N n^2
∞ 1
2. Denote Σ ─── is A
n=0 n!
1
Consider An=(1 + ──)^n
n
n n 1 n 1 n 1
= C + C *(──)^1 + ... + C *(──)^k + ... + C *(──)^n
0 1 n k n n n
So we have
n*(n-1)*...*(n-k+1) 1
= 1 + 1 + ... + ──────────*(──)^k + ...
k! n
where n*(n-1)*...*(n-k+1) 1 1 n n-1 n-k+1
──────────*(──)^k = ── * (──) * (──) *...*(───)
k! n k! n n n
1 1 k-1
= ── * 1 * (1- ──) *...*(1- ──)
k! n n
So we can rewrite An as:
1 1 1 1 2
An = 2 + ──*(1- ──) + ──*(1- ──)*(1- ──) +...
2! n 3! n n
1 1 k-1
+ ── * (1- ──) *...*(1- ──) + ...
k! n n
1 1 n-1
+ ── * (1- ──) *...*(1- ──)
n! n n
1
< e (Since each item is smaller then ──)
k!
So far we have showed that An is bounded
and An+1 > An is trivial (by comparing each item)
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 111.243.154.184
※ 编辑: znmkhxrw 来自: 111.243.154.184 (05/30 19:46)
1F:推 JohnMash :good 05/30 20:00