原出處在這裡
<題目>
P 27 5 P_{27}5 P275
- 若 ∑ a n \sum a_n ∑an收斂
- ∑ ( b n + 1 − b n ) \sum (b_{n+1}-b_n) ∑(bn+1−bn)絕對收斂
- 證明: ∑ a n b n 也 收 斂 \sum a_nb_n也收斂 ∑anbn也收斂
反思
- 确實想到了 ∣ ∑ k = n n + p a k b k ∣ = ∣ ∑ k = n n + p − 1 ( b k − b k + 1 ) S k + b n + p S n + p ∣ |\sum\limits_{k=n}^{n+p} a_kb_k|=|\sum\limits_{k=n}^{n+p-1}(b_{k}-b_{k+1})S_k+b_{n+p}S_{n+p}| ∣k=n∑n+pakbk∣=∣k=n∑n+p−1(bk−bk+1)Sk+bn+pSn+p∣ (1) ≤ ∣ ∑ k = n n + p − 1 ( b k − b k + 1 ) S k ∣ + ∣ b n + p S n + p ∣ \le|\sum\limits_{k=n}^{n+p-1}(b_{k}-b_{k+1})S_k|+|b_{n+p}S_{n+p}|\tag{1} ≤∣k=n∑n+p−1(bk−bk+1)Sk∣+∣bn+pSn+p∣(1)
- 但後面我就開始騷操作了, ∑ a n \sum a_n ∑an不是收斂嗎,那就讓 ∣ S k ∣ ≤ M |S_k|\le M ∣Sk∣≤M, ∑ k = n n + p − 1 ∣ b k − b k + 1 ∣ ≤ ε 2 M \sum\limits_{k=n}^{n+p-1}|b_{k}-b_{k+1}|\le\frac{\varepsilon}{2M} k=n∑n+p−1∣bk−bk+1∣≤2Mε,是以以上就有 ( 1 ) ≤ ε 2 + M ∣ b n + p ∣ (1)\le \frac{\varepsilon}2+M|b_{n+p}| (1)≤2ε+M∣bn+p∣
- 然後就難倒我了,這這這 ∣ b n + p ∣ |b_{n+p}| ∣bn+p∣咋求??在我苦思冥想之後,硬是想到了一種方法,不知道對不對:可以由已知條件證明 { b n } \{b_n\} {bn}收斂,用柯西條件:
- ∀ ε > 0 , ∃ N , 當 n − 1 > m > N 時 , 有 \forall\varepsilon>0,\exist N,當n-1>m>N時,有 ∀ε>0,∃N,當n−1>m>N時,有 ∣ b n − b m ∣ = ∣ T n − 1 − T m ∣ < ε |b_n-b_m|=|T_{n-1}-T_{m}|<\varepsilon ∣bn−bm∣=∣Tn−1−Tm∣<ε T n = ∑ i = 1 n ( b i + 1 − b i ) T_n=\sum\limits_{i=1}^n(b_{i+1}-b_i) Tn=i=1∑n(bi+1−bi)
- 是以 lim n → ∞ b n = 0 \lim\limits_{n\to\infty}b_n=0 n→∞limbn=0