※ 引述《preed (Freedom)》之銘言： : https://imgur.com/sbWYg1Z : 思考了一段時間 覺得還是超出自己能力範圍 : 可能還是需要數學高手協助或者給予提示 : 希望有版上數學好手支援 謝謝 (本篇因為有大量運算排版，建議手機使用者橫向閱讀) 這裡只證明第一小題，後面幾小題看了一下應該都滿簡單的（？） Let s_n = a_n + s, for n = 0, 1, 2, ... . -------(*) The original statement is equivalent to proving \lim_{n \to \infty} (a_0 + a_1 + ... + a_n)/(n+1) = 0 provided that \lim_{n \to \infty} a_n = 0. Assume \lim_{n \to \infty} a_n = 0. Since \lim_{n \to \infty} a_n = 0, given any \epsilon > 0, there exists a positive integer N_1 s.t. for all n > N_1 |a_n| < \epsilon/2. -------(A) For such fixed N_1, we can find another positive integer N_2 > N_1 s.t. |a_0 + a_1 + ... + a_{N_1}|/(n+1) < \epsilon/2. -------(B) In other words, given any \epsilon > 0, for n > N_2 (a_0 + a_1 + ... + a_n)/(n+1) = (a_0 + a_1 + ... + a_{N_1})/(n+1) + (a_{N_1+1} + a_{N_1+2} + ... + a_n)/(n+1) <= |a_0 + a_1 + ... + a_{N_1}|/(n+1) + (|a_{N_1+1}| + |a_{N_1+2}| + ... + |a_n|)/(n+1) < \epsilon/2 + (n-N_1)/(n+1) \epsilon/2 ^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^ from condition (B) from condition (A) < \epsilon/2 + \epsilon/2 = \epsilon. We've established if \lim_{n \to \infty} a_n = 0, then \lim_{n \to \infty} (a_0 + a_1 + ... + a_n)/(n+1) = 0, giving our desired result that if \lim_{n \to \infty} s_n = s (by (*)), \lim_{n \to \infty} (s_0 + s_1 + ... + s_n)/(n+1) = s. 有任何問題的話，歡迎有心的網友們一起來討論^^ --

※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 60.250.230.253 (臺灣) ※ 文章網址: https://webptt.com/m.aspx?n=bbs/Math/M.1588941479.A.80C.html ※ 編輯: cuylerLin (60.250.230.253 臺灣), 05/08/2020 20:42:40