Re: [分析] 級數收斂
※ 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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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
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