看板 Math 關於我們 聯絡資訊
※ 引述《eric80520 (freejustice)》之銘言: : 題目 Suppose that {ak} is a decreasing sequence of real numbers. : ∞ : Prove that if Σak converges, then kak→0 as k→∞. : k=1 : 我目前會的 : ∞ : since Σak converges,so ak→0 as k→∞ : k=1 : by {ak} is decreasing and →0 ,so {ak} is nonegative sequence : 再來好像要分2k跟2k+1討論, 可是我不太清楚要怎麼做 : 可以教我嗎 謝謝 Since Σa converges, then a → 0 as k → ∞ k k and since {a_k} is decreasing, then a ≧ 0 , for all k in N k hence ka = a + a + ... + a 2k 2k 2k 2k ≦ a + a + ... + a k+1 k+2 2k 2k = Σ a j=k+1 j ∵ Σa converges k 2k => |ka | ≦ | Σ a | → 0 as k → ∞ 2k j=k+1 j ∴ ka → 0 as k → ∞ 2k on the other hand 0 ≦ (2k+1)a ≦ (2k+1)a = 2ka + a → 0 as k → ∞ 2k+1 2k 2k 2k => (2k+1)a → 0 as k → ∞ 2k+1 therefore, ka → 0 as k → ∞ k -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 115.43.192.87
smartlwj :慢了一點,yhliu大大先po了 04/09 23:52
eric80520 :還是謝謝你們^_^ 04/10 10:01