Re: [分析] 初微(60)

看板Math作者 ( )時間20年前 (2005/08/21 09:12), 編輯推噓0(001)
留言1則, 1人參與, 最新討論串2/3 (看更多)
※ 引述《plover (>//////<)》之銘言: : If Σa_n converges with a_n > 0 for all n, and {a_n} is a : decreasing sequence, show that n a_n → 0 as n → +∞. By assumption => for any ε>0, there exists an integer N>0 s.t. n ≧ N we have a_(n+1) + ... + a_(n+n) < ε/2 {a_n} is a decreasing seq. => n[a_(2n)] < a_(n+1) + ... + a_(n+n) < ε/2 Hence 2n[a_(2n)] < ε whenever n ≧ N Similarly, a_(n+1) + ... + a_(n+n+1) < ε/2 whenever n ≧ N' for some N' then (n+1)a_(2n+1) < a_(n+1) + ... + a_(n+n+1) < ε/2 so (2n+1)a_(2n+1) < (2n+2)a_(2n+1) < ε Hence n(a_n) < ε whenever n ≧ max{N,N'} -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.219.178.211

61.219.178.211 08/21, , 1F
最後一行錯了 n 要調大一點 QQ
61.219.178.211 08/21, 1F
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