数列bn=2^n/(4^n-1),证明b1+b2+b3+……+bn_数学解答

数列bn=2^n/(4^n-1),证明b1+b2+b3+……+bn

人气：224 ℃　时间：2019-11-14 01:48:45

解答

bn=2^n/(4^n-1)b1= 2/3b2 = 4/15b3 = 8/63for n>=4bn =2^n/(4^n-1)< (2^n +1)/(4^n -1)= (2^n +1)/[(2^n -1)(2^n +1)]= 1/(2^n -1)< 1/2^(n-1)Sn = b1+b2+...+bn= 2/3 +4/15+8/63+ (b3+b4+...+bn)< 2/3 +4/15 +8/63+...