Re: [微積] 數列要如何證明收斂?

看板Math作者Desperato (Farewell)時間8年前 (2017/04/19 15:26)推噓5(5推 0噓 11→)

留言16則, 4人參與討論串2/2 (看更多)

※ 引述《shaqgod (ddinin)》之銘言： : 請問各位大大3.4題要如何證明收斂 : http://m.imgur.com/gallery/RmLoD0X : ----- : Sent from JPTT on my Asus ASUS_T00N. 3. Let z_0 = y_0 = a z_1 = y_1 - y_0 = b-a z_n = y_n - y_(n-1), n positive integer -1 Then z_n = --- z_(n-1) 3 -1 = (---)^(n-1) z_1 (z_0 is not in recursive formula) 3 n n -1 y_n = sum z_k = z_0 + z_1 sum (---)^(k-1) k=0 k=1 3 -1 1 3 Since the series {(---)^(n-1)} converges to -------- = --- 3 1-(-1/3) 4 the sequence {y_n} converges to a + (3/4)(b-a) = (a+3b)/4 4. Clearly x_n <= x_(n-1) for n >= 2 and x_n all positive thus x_n >= 0 for all n By Monotone sequence theorem {x_n} converges to some real number y_n >= y_(n-1) for n >= 2 We now try to find an upper bound M such that y_n <= M for all n Let 1 >= k > 0 M If there is an M such that ------- > 1 + k y_(n-1) M 1 + k 2^n 2^n - 1 Then --- = --------- = (1+k)------- > (1+k)------- y_n 1 + 1/2^n 2^n + 1 2^n 1 1+k 2 = (1+k)(1 - ---) = 1 + k - ----- >= 1 + k - --- 2^n 2^n 2^n 2 k 1 Now we want k - --- = ---, k = ------- 2^n 2 2^(n-2) M 1 Also ----- > 1 + --- = 2, take M = 4 y_1 2^0 Then above is exactly the induction proof M 1 for ----- > 1 + ------- for all n y_n 2^(n-1) Thus M > y_n for all n, M is an upper bound for {y_n} By Monotone sequence theorem {y_n} converges to some real number -- 嗯嗯ow o -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.25.105 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1492586764.A.E2F.html

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shaqgod

04/19 15:37, , 1^F

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yyc2008

04/19 15:40, , 2^F

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yyc2008

04/19 15:40, , 3^F

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