Re: 極限e的序列
※ 引述《LimSinE (r=e^theta)》之銘言:
: 熟知:
: (1+1/n)^n ↑e,(1+1/n)^(n+1)↓e (*)
: 事實上有
: (1+1/n)^(n+1/2)↓e (重點是遞減)
: 證明:
: 考慮函數 f(x) = (x+1/2) log (1+ 1/x)
: f'(x)
: = log (1 +1/x) - (x+1/2)/[x(x+1)]
: = log (1 +1/x) - 1/2 {1/x + 1/(x+1)} → 0 as x→無限大
: f"(x)
: = -1/[x(x+1)] + 1/2{1/x^2 + 1/(x+1)^2}
: =1/2{1/x - 1/(x+1)}^2 > 0,故f'↑0,f'<=0,f↓
: 請問有沒有不用微積分的方法?
: (例如(*)的兩個序列的增減都可用算幾不等式得到)
有點久的文章...
(1+1/n)^{n+1/2}>(1+1/{n+1})^{n+3/2}
iff (1+1/n)^{2n+1}>(1+1/{n+1})^{2n+3}
iff (n+1)^{4n+4}>n^{2n+1}*(n+2)^{2n+3}
iff (1+1/{n(n+2)})^{2n+2} > 1+2/n ~ (!)
Since
LHS of (!)
> 1+(2n+2)/{n(n+2)}+(2n+2)(2n+1)/{2n^2*(n+2)^2}
+(2n+2)(2n+1)(2n)/{6n^3*(n+2)^3}
= 1+2/n+{n^2+3n+8}/{3n^2*(n+2)^3}
> RHS of (!)
The first ineq holds!
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 1.160.224.47
※ 編輯: XII 來自: 1.160.224.47 (02/13 01:59)
推
02/13 11:21, , 1F
02/13 11:21, 1F
※ 編輯: XII 來自: 140.115.31.174 (02/13 14:02)
討論串 (同標題文章)