推 sm008150204 :>= |sin(π/2-1)|/|[π/2]|+ ... 這行不太懂 03/13 16:50
|sin([(2n+1)π/2])| >= |sin((2n+1)π/2-1)| = |sin(π/2-1)|
since [(2n+1)π/2] >= (2n+1)π/2-1
注意|sin(x)|在 (2n+1)π/2 >= x >= 2nπ/2 是遞增的且(2n+1)π/2-1 > 2nπ/2
※ 編輯: bineapple 來自: 133.3.254.116 (03/13 17:34)
※ 引述《zolAIsm3 (二胖子)》之銘言:
: ∞
: 證明Σ |sin(n)| / n 發散
: n=1
: 如果可以的話
: 不要使用Dirichlet Test
: 謝謝
以下的[ ]是高斯符號(floor)
∞
Σ |sin(n)|/n > |sin([π/2])|/|[π/2]| + |sin([3π/2])|/|[3π/2]|
n=1
+ |sin([5π/2])|/|[5π/2]| + .....
>= |sin(π/2-1)|/|[π/2]| + |sin(π/2-1)|/|[3π/2]|
+ |sin(π/2-1)|/|[5π/2]| + .....
(設|sin(π/2-1)|=C>0)
>= C * ( 2/π + 2/(3π) + 2/(5π) + .... )
= 2C/π * (1+1/3+1/5+...)
> 2C/π * (1/2+1/4+1/6+...)
= C/π * (1+1/2+1/3+...) = ∞
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◆ From: 126.61.45.199
※ 編輯: bineapple 來自: 126.61.45.199 (03/13 08:54)
※ 編輯: bineapple 來自: 126.61.45.199 (03/13 09:11)