题目

已知不等式++…+＞［log2n］,其中n为大于2的整数,［log2n］表示不超过log2n的最大整数,设数列{an}的各项为正,且满足a1=b(b＞0),an≤,n=2,3,4,…(1)证明an＜,n=3,4,5,…，(2)猜测数列{an}是否有极限?如果有,写出极限的值(不必证明)；(3)试确定一个正整数N,使得当n＞N时,对任意b＞0,都有an＜. 答案：(1)证法1：∵当n≥2时,0＜an≤,∴≥=+,即-≥.于是有-≥,-≥,…,-≥.所有不等式两边相加可得-≥++…+.由已知不等式知,当n≥3时有,-＞［log2n］.∵a1=b,∴＞+［log2n］=.∴an＜，n=3,4,5….证法2:设f(n)=++…+,首先利用数学归纳法证不等式an≤,n=3,4,5,….①当n=3时,由a3≤=≤=,知不等式成立.②假设当n=k(k≥3)时,不等式--Thanks a lot for helping me.--________.A.You are welcomeB.It doesn't matterC.No, thank youD.It's kind of you