推 superpigpig:謝謝，我看懂了，只是第一小題中間好像有幾行怪怪的? 04/13 10:21
p , q 為二正整數， p/q = 1/(1‧2)+1/(3‧4)+1/(5‧6)+…+1/(99‧100)
試證 (A) p 為 151 的倍數
(B) 0 < ln2-p/q < 0.01
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作者: Dirichlet (微風輕吹) 看板: tutor
標題: Re: [解題]倍數相關的証明題
時間: Thu Apr 13 10:07:55 2006
※ 引述《superpigpig (豬豬)》之銘言：
: p , q 為二正整數， p/q = 1/(1‧2)+1/(3‧4)+1/(5‧6)+…+1/(99‧100)
: 試證 (A) p 為 151 的倍數
: (B) 0 < ln2-p/q < 0.01
1/(1‧2) + 1/(3‧4) + 1/(5‧6) + … + 1/(99‧100)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/99 - 1/100
= (1 + 1/3 + ... + 1/99) - (1/2)(1 + 1/2 + ... + 1/50)
= (1 + 1/2 + ... + 1/100) - (1 + 1/2 + ... + 1/50)
= 1/51 + 1/52 + ... + 1/100
= 151‧[1/(100‧51) + 1/(99‧52) + ... + 1/(76‧75)]
Hence p = 151‧[qM/(51‧...‧100)] for some integer M>0,
note that (151 , 51‧...‧100) = 1 and qM/(51‧...‧100) is a integer,
so 151 | p
oo 1
ln2 - p/q = Σ -------------- > 0
n=50 (2n+1)(2n+2)
oo 1
ln2 - p/q = 1/101 - Σ -------------- < 1/101 < 1/100 = 0.01
n=50 (2n+2)(2n+3)
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◆ From: 140.112.217.78
※ 編輯: Dirichlet 來自: 140.112.217.78 (04/13 10:09)