Re: [分析] 級數收斂

[Desperato (Farewell)]

※ 引述《BrowningZen (BrowningZen)》之銘言：
: a我硬是用integral test避開了問題,請問有其他方法嗎?
: 請問b要怎麼做呢,除了把整條式子變成a的形式後就沒有頭緒了
: 謝謝各位
: http://i.imgur.com/b3vpWTg.jpg


(a) if k > 1 then

    1     1     1     1           1     1
   --- + --- + --- + --- + ... + --- + --- + ...
   1^k   2^k   3^k   4^k         7^k   8^k

    1     1     1     1           1     1
 < --- + --- + --- + --- + ... + --- + --- + ... converges
   1^k   2^k   2^k   4^k         4^k   8^k

可以查Cauchy condensation test



(b) Let t = beta/2

There exists some natural number N0 such that

                 |b_(k+1)|
k > N0 implies k(--------- - 1) < - 1 - t
                   |b_k|

                            1+t
Then |b_(k+1)| < |b_k| (1 - ---)
                             k

                   M
Now let a_k = -----------
              (k-1)^(1+t)

     a_(k+1)    k-1               1+t
Then ------- = (---)^(1+t) > (1 - ---)
       a_k       k                 k

(by binomial theorem and alternating series)


Then |b_k| < a_k implies

                     1+t
|b_(k+1)| < a_k (1 - ---) < a_(k+1)
                      k


choose a_1 > |b_1| and some a_2 = M > |b_2|

Then above gives an induction prove for |b_k| < a_k for all k


sum a_k converges by (a)


--

嗯嗯ow o

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※ 文章網址: https://www.ptt.cc/bbs/Math/M.1493909656.A.F08.html
→ Desperato   : 啊啊不對 k要大於N0才行 反正改一改就好
推 BrowningZen : 謝謝大大 真的完全想不到!
→ yyc2008     : k > N0那一行 怎麼知道是小於? 不等號左邊是不是要
→ yyc2008     : 加大絕對值框住左邊所有項?
Let t = beta/2

There exists some natural number N0 such that

               |   |b_(k+1)|                   |
k > N0 implies | k(--------- - 1) - (- 1 - 2t) | < t
               |     |b_k|                     |

拆絕對值 把(-1-2t)往右丟過去

※ 編輯: Desperato (140.112.25.105), 05/06/2017 03:09:24
→ yyc2008     : 謝謝  看懂了