证明: ① an = Sn - Sn-1 an = a/(1-a) * (1 - an ) - a/(1-a) * (1 - an-1) 两边同乘(1-a) (1-a) an = a(1 - an ) - a(1 - an-1) an - a * an = a - a * an - a + a * an-1 an = a * an-1 an:an-1=a 由于常数 a != 1,所以{an}是等比数列. ② bn=an lg an 根据第一题的证明可知bn是: (a lg a)、(a^2 lg a^2)...(a^n lg a^n) 根据题意:对于任何正整数n具有bn>=bm 所以: (a^n lg a^n ) >= a^m lg a^m 即: n*a^n log a >= m*a^m log a 由于log a = log √7/3 < 0 所以: n a^n
an = Sn - Sn-1 = a/(1-a) * (1 - an ) - a/(1-a) * (1 - an-1) 两边同乘(1-a) (1-a) an = a(1 - an ) - a(1 - an-1) an - a * an = a - a * an - a + a * an-1 an = a * an-1 an:an-1=a ...