已知: ∀ ε > 0 , ∃ n > N , ∣ a n − A ∣ < ε \forall \varepsilon >0, \exist n>N,|a_n-A|<\varepsilon ∀ε>0,∃n>N,∣an−A∣<ε
目标: ∀ ε > 0 , ∃ n > N 1 , ∣ a 1 + . . . + a n n − A ∣ < ε \forall \varepsilon >0, \exist n>N_1,|\frac{a_1+...+a_n}{n}-A|<\varepsilon ∀ε>0,∃n>N1,∣na1+...+an−A∣<ε
假设 S n = a 1 + . . . + a n S_n=a_1+...+a_n Sn=a1+...+an,则变成了求 ∣ S n n − A ∣ < ε |\frac{S_n}n-A|<\varepsilon ∣nSn−A∣<ε
由已知条件, n > N , ∣ a n − A ∣ < ε n>N,|a_n-A|<\varepsilon n>N,∣an−A∣<ε,把n个数分为小于等于N的和大于N的分别处理, ∣ S n n − A ∣ < ∣ S N + N ∣ A ∣ + ∣ a N + 1 − A ∣ + . . . + ∣ a n − A ∣ n ∣ < ∣ S N + N ∣ A ∣ + ( n − N ) ε n ∣ < ∣ S N + N ∣ A ∣ + ε n ∣ |\frac{S_n}n-A|<|\frac{S_N+N|A|+|a_{N+1}-A|+...+|a_n-A|}n|<|\frac{S_N+N|A|+(n-N)\varepsilon}n|<|\frac{S_N+N|A|+\varepsilon}n| ∣nSn−A∣<∣nSN+N∣A∣+∣aN+1−A∣+...+∣an−A∣∣<∣nSN+N∣A∣+(n−N)ε∣<∣nSN+N∣A∣+ε∣
因为 S N , N ∣ A ∣ S_N,N|A| SN,N∣A∣是常数,所以 l i m n − > ∞ S N + N ∣ A ∣ n = 0 lim_{n->\infin}\frac{S_N+N|A|}n=0 limn−>∞nSN+N∣A∣=0,所以当 N 1 > N N_1>N N1>N时 l i m n − > ∞ ∣ S n n − A ∣ < ε lim_{n->\infin}|\frac{S_n}n-A|<\varepsilon limn−>∞∣nSn−A∣<ε