=(1/(k+1)+1/(k+2)+1/(k+3)+"..." +1/(3k+1))+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)>25/24+[1/(3k+2)+1/(3k+4) "-" 2/(3"(" k+1")" )].

　　因为1/(3k+2)+1/(3k+4)=(6"(" k+1")" )/(9k^2+18k+8)>(6"(" k+1")" )/(9k^2+18k+9)=(6"(" k+1")" )/(9"(" k+1")" ^2 )=2/(3"(" k+1")" ),

　　所以1/(3k+2)+1/(3k+4)-2/(3"(" k+1")" )>0,

　　于是1/("(" k+1")" +1)+1/("(" k+1")" +2)+1/("(" k+1")" +3)+...+1/(3"(" k+1")" +1)>25/24,

　　即当n=k+1时不等式成立.

　　由(1)(2)知,对一切正整数n,都有1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24,且正整数a的最大值等于25.

10.导学号26394069已知数列{an}满足:a1=3/2,且an=(3na_(n"-" 1))/(2a_(n"-" 1)+n"-" 1)(n≥2,n∈N+).

(1)求数列{an}的通项公式;

(2)求证对一切正整数n,不等式a1a2...an<2n!恒成立.

(1)解将条件变为1-n/a_n =1/3 (1"-" (n"-" 1)/a_(n"-" 1) ),

　　因此数列{1"-" n/a_n }为一个等比数列,其首项为1-1/a_1 =1/3,公比为1/3,

　　从而1-n/a_n =1/3^n ,

　　因此得an=(n×3^n)/(3^n "-" 1)(n≥1).0①

(2)证明由①得

　　a1a2...an=n"!" /((1"-" 1/3)×(1"-" 1/3^2 )×"..." ×(1"-" 1/3^n ) ).

　　为证a1a2...an<2n!,只要证当n∈N+时,有(1"-" 1/3)×(1"-" 1/3^2 )×...×(1"-" 1/3^n )>1/2.0②

　　显然,左端每个因式皆为正数,先证明对n∈N+,有

　　(1"-" 1/3)×(1"-" 1/3^2 )×...×(1"-" 1/3^n )

　　≥1-(1/3+1/3^2 +"..." +1/3^n ).0③

　　下面用数学归纳法证明③式:

　　ⅰ当n=1时,显然③式成立,

　　ⅱ假设当n=k(k≥1)时,③式成立,

　　即(1"-" 1/3)×(1"-" 1/3^2 )×...×(1"-" 1/3^k )≥

1-(1/3+1/3^2 +"..." +1/3^k ).