Mertens定理（级数乘法）
【定理3.50】
对于复数序列 { a n } 、 { b n } 、 { c n } \left\{ a_{n} \right\} 、\left\{ b_{n} \right\} 、\left\{ c_{n} \right\} {an}、{bn}、{cn}，假定
a. ∑ n = 0 ∞ a n \sum_{n = 0}^{\infty}a_{n} ∑n=0∞an绝对收敛
b. ∑ n = 0 ∞ a n = A \sum_{n = 0}^{\infty}a_{n} = A ∑n=0∞an=A
c. ∑ n = 0 ∞ b n = B \sum_{n = 0}^{\infty}b_{n} = B ∑n=0∞bn=B
d. c n = ∑ k = 0 n a k b n − k ( n = 0 , 1 , 2 , … ) c_{n} = \sum_{k = 0}^{n}{a_{k}b_{n - k}}\ \ \ (n = 0,1,2,\ldots) cn=∑k=0nakbn−k (n=0,1,2,…)
那么
∑ n = 0 ∞ c n = A B \sum_{n = 0}^{\infty}c_{n} = AB n=0∑∞cn=AB
【证明】
设
A n = ∑ i = 0 n a i A_{n} = \sum_{i = 0}^{n}a_{i} An=i=0∑nai
B n = ∑ i = 0 n b i B_{n} = \sum_{i = 0}^{n}b_{i} Bn=i=0∑nbi
C n = ∑ i = 0 n c i C_{n} = \sum_{i = 0}^{n}c_{i} Cn=i=0∑nci
B ∗ = l i m d i a m n → ∞ B n B^{*} = \underset{n \rightarrow \infty}{limdiam}B_{n} B∗=n→∞limdiamBn
D = ∑ i = 1 ∞ ∣ a i ∣ D = \sum_{i = 1}^{\infty}\left| a_{i} \right| D=i=1∑∞∣ai∣
D n = ∑ i = 1 n ∣ a i ∣ D_{n} = \sum_{i = 1}^{n}\left| a_{i} \right| Dn=i=1∑n∣ai∣
若 D = 0 D = 0 D=0，则 a i = 0 a_{i} = 0 ai=0，结论显然成立，下面考虑更一般的情况，即 D > 0 D > 0 D>0
对于任意实数 ε > 0 \varepsilon > 0 ε>0，令
δ = min ( ε 3 D , ε 3 B ∗ , D 3 ) \delta = \min\left( \frac{\varepsilon}{3D},\frac{\varepsilon}{3B^{*}},\frac{D}{3} \right) δ=min(3Dε,3B∗ε,3D)
则
0 < δ < D 0 < \delta < D 0<δ<D
( D + δ ) δ + B ∗ δ < ε (D + \delta)\delta + B^{*}\delta < \varepsilon (D+δ)δ+B∗δ<ε
对于这样的 δ \delta δ，可以找到 N 1 , N 2 , N 3 N_{1},N_{2},N_{3} N1,N2,N3，使得
∀ n ≥ N 1 , 1 ≤ i ≤ n [ ∣ B 2 n − B 2 n − i ∣ < δ ] \forall n \geq N_{1},1 \leq i \leq n\left\lbrack \left| B_{2n} - B_{2n - i} \right| < \delta \right\rbrack ∀n≥N1,1≤i≤n[∣B2n−B2n−i∣<δ]
∀ n ≥ N 2 [ D n < δ + D ] \forall n \geq N_{2}\left\lbrack D_{n} < \delta + D \right\rbrack ∀n≥N2[Dn<δ+D]
∀ n ≥ N 3 [ ∣ D 2 n − D n ∣ < δ ] \forall n \geq N_{3}\left\lbrack \left| D_{2n} - D_{n} \right| < \delta \right\rbrack ∀n≥N3[∣D2n−Dn∣<δ]
取
N = max ( N 1 , N 2 , N 3 ) N = \max{(N_{1},N_{2},N_{3})} N=max(N1,N2,N3)
则对于任意的 n ≥ N n \geq N n≥N， 1 ≤ i ≤ n 1 \leq i \leq n 1≤i≤n都有
∣ B 2 n − B 2 n − i ∣ < δ \left| B_{2n} - B_{2n - i} \right| < \delta ∣B2n−B2n−i∣<δ
D n < δ + D D_{n} < \delta + D Dn<δ+D
∣ D 2 n − D n ∣ < δ \left| D_{2n} - D_{n} \right| < \delta ∣D2n−Dn∣<δ
设
E 2 n = A 2 n B 2 n − C 2 n E_{2n} = A_{2n}B_{2n} - C_{2n} E2n=A2nB2n−C2n
由于
E 2 n = ∑ i = 1 2 n ∑ j = 2 n − i + 1 2 n a i b i = ∑ i = 1 2 n ( B 2 n − B 2 n − i ) a i = ∑ i = 1 n ( B 2 n − B 2 n − i ) a i + ∑ i = 1 n ( B 2 n − B n − i ) a n + i E_{2n} = \sum_{i = 1}^{2n}{\sum_{j = 2n - i + 1}^{2n}{a_{i}b_{i}}} = \sum_{i = 1}^{2n}{\left( B_{2n} - B_{2n - i} \right)a_{i}} = \sum_{i = 1}^{n}{\left( B_{2n} - B_{2n - i} \right)a_{i}} + \sum_{i = 1}^{n}{\left( B_{2n} - B_{n - i} \right)a_{n + i}} E2n=i=1∑2nj=2n−i+1∑2naibi=i=1∑2n(B2n−B2n−i)ai=i=1∑n(B2n−B2n−i)ai+i=1∑n(B2n−Bn−i)an+i
则
∣ E 2 n ∣ ≤ ∑ i = 1 n ∣ B 2 n − B 2 n − i ∣ ∣ a i ∣ + ∑ i = 1 n ∣ B 2 n − B n − i ∣ ∣ a n + i ∣ \left| E_{2n} \right| \leq \sum_{i = 1}^{n}{\left| B_{2n} - B_{2n - i} \right|\left| a_{i} \right|} + \sum_{i = 1}^{n}{\left| B_{2n} - B_{n - i} \right|\left| a_{n + i} \right|} ∣E2n∣≤i=1∑n∣B2n−B2n−i∣∣ai∣+i=1∑n∣B2n−Bn−i∣∣an+i∣
若 n ≥ N n \geq N n≥N，可得
∣ E 2 n ∣ < δ ∑ i = 1 n ∣ a i ∣ + B ∗ ∑ i = 1 n ∣ a n + i ∣ = D n δ + B ∗ ∣ D 2 n − D n ∣ < ( D + δ ) δ + B ∗ δ < ε \left| E_{2n} \right| < \delta\sum_{i = 1}^{n}\left| a_{i} \right| + B^{*}\sum_{i = 1}^{n}\left| a_{n + i} \right| = D_{n}\delta + B^{*}\left| D_{2n} - D_{n} \right| < (D + \delta)\delta + B^{*}\delta < \varepsilon ∣E2n∣<δi=1∑n∣ai∣+B∗i=1∑n∣an+i∣=Dnδ+B∗∣D2n−Dn∣<(D+δ)δ+B∗δ<ε
也就是说，对于任意实数 ε > 0 \varepsilon > 0 ε>0，都有 N N N，使得 n ≥ N n \geq N n≥N时
∣ E 2 n ∣ = ∣ A 2 n B 2 n − C 2 n ∣ < ε \left| E_{2n} \right| = \left| A_{2n}B_{2n} - C_{2n} \right| < \varepsilon ∣E2n∣=∣A2nB2n−C2n∣<ε
类似地，可以得到
∣ E 2 n + 1 ∣ = ∣ A 2 n + 1 B 2 n + 1 − C 2 n + 1 ∣ < ε \left| E_{2n + 1} \right| = \left| A_{2n + 1}B_{2n + 1} - C_{2n + 1} \right| < \varepsilon ∣E2n+1∣=∣A2n+1B2n+1−C2n+1∣<ε
所以
lim n → ∞ C n = lim n → ∞ A n B n \lim_{n \rightarrow \infty}C_{n} = \lim_{n \rightarrow \infty}{A_{n}B_{n}} n→∞limCn=n→∞limAnBn
而
lim n → ∞ A n B n = lim n → ∞ A n × lim n → ∞ B n = A B \lim_{n \rightarrow \infty}{A_{n}B_{n}} = \lim_{n \rightarrow \infty}A_{n} \times \lim_{n \rightarrow \infty}B_{n} = AB n→∞limAnBn=n→∞limAn×n→∞limBn=AB
所以
∑ n = 0 ∞ c n = lim n → ∞ C n = A B \sum_{n = 0}^{\infty}c_{n} = \lim_{n \rightarrow \infty}C_{n} = AB n=0∑∞cn=n→∞limCn=AB