2很简单,试一下就出来了
3.若否,则存正整数N≥3,对n＞N,an＞0(n＞N≥3故an是绝对值差≥0)
此时在n＞N时，an+2=|an+1-an|＜max{an,an+1},同理有an+3＜max{an+2,an+1}＜max{max{an,an+1},an+1}=max{an,an+1},由此得max{an+3,an+2}＜max{an,an+1},计cn=max{an,an+1},则cn＞cn+2＞cn+4＞...无穷递降,而cn是正整数,cn＞0,矛盾！故数列中必有无穷多个0
3.若否,则存正整数N≥3,对n＞N,an＞0(n＞N≥3故an是绝对值差≥0)
此时在n＞N时，an+2=|an+1-an|＜max{an,an+1},同理有an+3＜max{an+2,an+1}＜max{max{an,an+1},an+1}=max{an,an+1},由此得max{an+3,an+2}＜max{an,an+1},计cn=max{an,an+1},则cn＞cn+2＞cn+4＞...无穷递降,而cn是正整数,cn＞0,矛盾！故数列中必有无穷多个0
