Toeplitz定理推广和应用
Toeplitz定理推广和应用
Toeplitz定理
设 n , k ∈ N n,k \in \mathbb{N} n,k∈N， t n k ≥ 0 t_{nk}\ge0 tnk≥0 且 ∑ k = 1 n t n k = 1 \sum_{k=1}^{n}t_{nk}=1 ∑k=1ntnk=1， lim n → + ∞ t n k = 0 \lim_{n\rightarrow +\infty}t_{nk}=0 limn→+∞tnk=0。如果 lim n → + ∞ a n = a \lim_{n \rightarrow + \infty}a_n=a limn→+∞an=a，则
lim n → + ∞ ∑ k = 1 n t n k ⋅ a k = a \lim_{n \rightarrow +\infty} \sum_{k=1}^{n}t_{nk} \cdot a_k=a n→+∞limk=1∑ntnk⋅ak=a
证略
推广 把和为1改为和的极限为1
设 n , k ∈ N n,k \in \mathbb{N} n,k∈N， t n k ≥ 0 t_{nk}\ge0 tnk≥0 且 lim n → + ∞ ∑ k = 1 n t n k = 1 \lim_{n → +\infty}\sum_{k=1}^{n}t_{nk}=1 limn→+∞∑k=1ntnk=1， lim n → + ∞ t n k = 0 \lim_{n\rightarrow +\infty}t_{nk}=0 limn→+∞tnk=0。如果 lim n → + ∞ a n = a \lim_{n \rightarrow + \infty}a_n=a limn→+∞an=a，则
lim n → + ∞ ∑ k = 1 n t n k ⋅ a k = a \lim_{n \rightarrow +\infty} \sum_{k=1}^{n}t_{nk} \cdot a_k=a n→+∞limk=1∑ntnk⋅ak=a
证明:
设 S n = ∑ k = 1 n t n k S_n=\sum_{k=1}^{n}{t_{nk}} Sn=∑k=1ntnk
因为 lim n → + ∞ t n k = 1 > 0 \lim_{n→ +\infty}t_{nk}=1 \gt 0 limn→+∞tnk=1>0,因此 ∃ N 1 ∈ N \exist N_1∈ \mathbb{N} ∃N1∈N,当 n > N 1 n \gt N_1 n>N1时， S n > 0 S_n \gt 0 Sn>0
当 n > N 1 n \gt N_1 n>N1时，令 b n k = t n k / S n b_{nk}=t_{nk}/S_n bnk=tnk/Sn，则
b n k ≥ 0 b_{nk} \ge 0 bnk≥0并且
∑ k = 1 n b n k = 1 \sum_{k=1}^n{b_{nk}=1} ∑k=1nbnk=1， lim n → + ∞ b n k = 0 \lim_{n→ +\infty}b_{nk}=0 limn→+∞bnk=0
根据Toeplitz定理
lim n → + ∞ ∑ k = 1 n b n k ⋅ a k = a \lim_{n \rightarrow +\infty} \sum_{k=1}^{n}b_{nk} \cdot a_k=a n→+∞limk=1∑nbnk⋅ak=a
所以
lim n → + ∞ ∑ k = 1 n t n k ⋅ a k = lim n → + ∞ S n ⋅ ∑ k = 1 n b n k ⋅ a k = a \lim_{n \rightarrow +\infty} \sum_{k=1}^{n}t_{nk} \cdot a_k= \lim_{n \rightarrow +\infty} S_n ⋅ \sum_{k=1}^{n}b_{nk} \cdot a_k = a n→+∞limk=1∑ntnk⋅ak=n→+∞limSn⋅k=1∑nbnk⋅ak=a
定理的应用
1. 设 lim n → + ∞ a n = a \lim_{n\rightarrow+\infty} a_n=a limn→+∞an=a，证明
lim n → + ∞ a 1 + 2 a 2 + ⋯ + n a n n 2 = a 2 \lim_{n\rightarrow+\infty} \frac{a_1+2a_2+\cdots+na_n}{n^2}=\frac{a}{2} n→+∞limn2a1+2a2+⋯+nan=2a
证明：
设 t n k = 2 k n 2 t_{nk}=\frac{2k}{n^2} tnk=n22k，易知 t n k ≥ 0 t_{nk} \ge 0 tnk≥0， lim n → + ∞ ∑ k = 1 n t n k = 1 \lim_{n\rightarrow+\infty}\sum_{k=1}^{n}t_{nk}=1 limn→+∞∑k=1ntnk=1， lim n → + ∞ t n k = 0 \lim_{n\rightarrow+\infty}t_{nk}=0 limn→+∞tnk=0
故由Toeplitz定理的推广知，
lim n → + ∞ t n k ⋅ a k = a \lim_{n\rightarrow+\infty} t_{nk} ⋅ a_k=a n→+∞limtnk⋅ak=a
即
lim n → + ∞ 2 ⋅ a 1 + 2 a 2 + ⋯ + n a n n 2 = a \lim_{n\rightarrow+\infty} 2⋅ \frac{a_1+2a_2+\cdots+na_n}{n^2}=a n→+∞lim2⋅n2a1+2a2+⋯+nan=a
左边乘以 1 2 \frac{1}{2} 21,即得
lim n → + ∞ a 1 + 2 a 2 + ⋯ + n a n n 2 = a 2 \lim_{n\rightarrow+\infty} \frac{a_1+2a_2+\cdots+na_n}{n^2}=\frac{a}{2} n→+∞limn2a1+2a2+⋯+nan=2a
2. 设 lim n → + ∞ a n = a \lim_{n\rightarrow+\infty} a_n=a limn→+∞an=a， 0 < q < 1 0<q<1 0<q<1 证明
lim n → + ∞ ( a n + a n − 1 ⋅ q + ⋯ + a 1 ⋅ q n − 1 ) = a 1 − q \lim_{n\rightarrow+\infty} \left( a_n + a_{n-1} ⋅ q+ \cdots + a_1 ⋅ q^{n-1} \right) = \frac{a}{1-q} n→+∞lim(an+an−1⋅q+⋯+a1⋅qn−1)=1−qa
证明：
设 t n k = ( 1 − q ) ⋅ q n − k t_{nk}=(1-q)⋅ q^{n-k} tnk=(1−q)⋅qn−k，易知 t n k ≥ 0 t_{nk} \ge 0 tnk≥0， lim n → + ∞ ∑ k = 1 n t n k = 1 \lim_{n\rightarrow+\infty}\sum_{k=1}^{n}t_{nk}=1 limn→+∞∑k=1ntnk=1， lim n → + ∞ t n k = 0 \lim_{n\rightarrow+\infty}t_{nk}=0 limn→+∞tnk=0
故由Toeplitz定理的推广知，
lim n → + ∞ t n k ⋅ a k = a \lim_{n\rightarrow+\infty} t_{nk} ⋅ a_k=a n→+∞limtnk⋅ak=a
即
lim n → + ∞ ( 1 − q ) ( a 1 ⋅ q n − 1 + ⋯ + a n − 1 ⋅ q + a n ) = a \lim_{n\rightarrow+\infty} \left(1-q\right) \left( a_1⋅ q^{n-1} + \cdots + a_{n-1} ⋅ q + a_n \right)=a n→+∞lim(1−q)(a1⋅qn−1+⋯+an−1⋅q+an)=a
左边除以 1 − q 1-q 1−q,即得
lim n → + ∞ ( a n + a n − 1 ⋅ q + ⋯ + a 1 ⋅ q n − 1 ) = a 1 − q \lim_{n\rightarrow+\infty} \left( a_n + a_{n-1} ⋅ q+ \cdots + a_1 ⋅ q^{n-1} \right) = \frac{a}{1-q} n→+∞lim(an+an−1⋅q+⋯+a1⋅qn−1)=1−qa
3. 设 lim n → + ∞ a n = a \lim_{n\rightarrow+\infty} a_n=a limn→+∞an=a， lim n → + ∞ b n = b > 0 \lim_{n\rightarrow+\infty} b_n=b > 0 limn→+∞bn=b>0，且 b n > 0 , ∀ n ∈ N b_n \gt 0, ∀ n ∈ \mathbb{N} bn>0,∀n∈N,证明
lim n → + ∞ a 1 b n + a 2 b n − 1 + ⋯ + a n b 1 n = a b \lim_{n\rightarrow+\infty} \frac{a_1b_n+a_2b_{n-1} + \cdots + a_nb_1}{n}=ab n→+∞limna1bn+a2bn−1+⋯+anb1=ab
证明：
设 t n k = b n − k + 1 n ⋅ b t_{nk}=\frac{b_{n-k+1}}{n ⋅ b} tnk=n⋅bbn−k+1，易知 t n k ≥ 0 t_{nk} \ge 0 tnk≥0， lim n → + ∞ t n k = 0 \lim_{n\rightarrow+\infty}t_{nk}=0 limn→+∞tnk=0
而 lim n → + ∞ ∑ k = 1 n t n k = lim n → + ∞ 1 b ⋅ b 1 + b 2 + ⋯ + b n n = 1 b ⋅ lim n → + ∞ b n = 1 \lim_{n\rightarrow+\infty}\sum_{k=1}^{n}t_{nk}=\lim_{n\rightarrow+\infty} \frac{1}{b}⋅ \frac{b_1+b_2+\cdots+b_n}{n}=\frac{1}{b}⋅ \lim_{n\rightarrow+\infty}b_n=1 limn→+∞∑k=1ntnk=limn→+∞b1⋅nb1+b2+⋯+bn=b1⋅limn→+∞bn=1
故由Toeplitz定理的推广知，
lim n → + ∞ t n k ⋅ a k = a \lim_{n\rightarrow+\infty} t_{nk} ⋅ a_k=a n→+∞limtnk⋅ak=a
即
lim n → + ∞ b n n ⋅ b ⋅ a 1 + b n − 1 n ⋅ b ⋅ a 2 + b 1 n ⋅ b ⋅ a n = lim n → + ∞ a 1 ⋅ b n + a 2 ⋅ b n − 1 ⋯ + a n ⋅ b 1 n ⋅ b = a \lim_{n\rightarrow+\infty} \frac{b_n}{n⋅ b}⋅ a_1 + \frac{b_{n-1}}{n⋅ b}⋅ a_2 + \frac{b_1}{n⋅ b}⋅ a_n=\lim_{n\rightarrow+\infty} \frac{a_1⋅ b_n + a_2⋅ b_{n-1} \cdots + a_n⋅ b_1}{n⋅ b}=a n→+∞limn⋅bbn⋅a1+n⋅bbn−1⋅a2+n⋅bb1⋅an=n→+∞limn⋅ba1⋅bn+a2⋅bn−1⋯+an⋅b1=a
两边同时乘以 b b b，即得
lim n → + ∞ a 1 b n + a 2 b n − 1 + ⋯ + a n b 1 n = a b \lim_{n\rightarrow+\infty} \frac{a_1b_n+a_2b_{n-1} + \cdots + a_nb_1}{n}=ab n→+∞limna1bn+a2bn−1+⋯+anb1=ab
4. 设 lim n → + ∞ a n = a \lim_{n\rightarrow+\infty}a_n=a limn→+∞an=a， b n ≥ 0 ( n ∈ N ) b_n \ge 0 (n∈ \mathbb{N}) bn≥0(n∈N)， lim n → + ∞ ( b 1 + b 2 + ⋯ + b n ) = S \lim_{n\rightarrow+\infty} \left( b_1 +b_2 + \cdots + b_n\right)=S limn→+∞(b1+b2+⋯+bn)=S，证明:
lim n → + ∞ ( a 1 ⋅ b n + a 2 ⋅ b n − 1 + ⋯ + a n ⋅ b 1 ) = a ⋅ S \lim_{n\rightarrow+\infty} \left( a_1⋅ b_n + a_2⋅ b_{n-1} + \cdots + a_n⋅ b_1 \right)=a⋅ S n→+∞lim(a1⋅bn+a2⋅bn−1+⋯+an⋅b1)=a⋅S
证明：
情况1. S = 0 S=0 S=0,则 b n = 0 b_n=0 bn=0,题目得证
情况2. S > 0 S\gt 0 S>0，令 S n = b 1 + b 2 + ⋯ + b n S_n=b_1+b_2+\cdots+b_n Sn=b1+b2+⋯+bn， b n = S n − S n − 1 b_n=S_n-S_{n-1} bn=Sn−Sn−1,则 lim n → + ∞ b n = lim n → + ∞ ( S n − S n − 1 ) = S − S = 0 \lim_{n\rightarrow+\infty}b_n=\lim_{n\rightarrow+\infty}\left( S_n - S_{n-1}\right)=S-S=0 limn→+∞bn=limn→+∞(Sn−Sn−1)=S−S=0
令 t n k = b n − k + 1 S t_{nk}=\frac{b_{n-k+1}}{S} tnk=Sbn−k+1 易知 t n k ≥ 0 t_{nk} \ge 0 tnk≥0， lim n → + ∞ ∑ k = 1 n t n k = 1 \lim_{n\rightarrow+\infty}\sum_{k=1}^{n}t_{nk}=1 limn→+∞∑k=1ntnk=1， lim n → + ∞ t n k = 0 \lim_{n\rightarrow+\infty}t_{nk}=0 limn→+∞tnk=0
故由Toeplitz定理的推广知，
lim n → + ∞ t n k ⋅ a k = a \lim_{n\rightarrow+\infty} t_{nk} ⋅ a_k=a n→+∞limtnk⋅ak=a
即
lim n → + ∞ a 1 ⋅ b n + a 2 ⋅ b n − 1 + ⋯ + a n ⋅ b 1 S = a \lim_{n\rightarrow+\infty} \frac{a_1⋅ b_n + a_2⋅ b_{n-1} + \cdots + a_n⋅ b_1}{S}=a n→+∞limSa1⋅bn+a2⋅bn−1+⋯+an⋅b1=a
两边同乘以 S S S,即得
lim n → + ∞ ( a 1 ⋅ b n + a 2 ⋅ b n − 1 + ⋯ + a n ⋅ b 1 ) = a ⋅ S \lim_{n\rightarrow+\infty} \left( a_1⋅ b_n + a_2⋅ b_{n-1} + \cdots + a_n⋅ b_1 \right)=a⋅ S n→+∞lim(a1⋅bn+a2⋅bn−1+⋯+an⋅b1)=a⋅S