已知函数f（x）=（3x+2）/（x+2）,（1）若数列﹛an﹜,﹛bn﹜满足a1=1/2,a(n+1)＜注：下脚标＞f

(1)b1=1/(a1+1)=1/(1/2+1)=2/3b(n+1)=1/(f(an)+1)=(an+2)/(4an+4)=1/4(1+1/(an+1))=1/4(1+bn)==>bn=1/4+1/4b(n-1)=1/4+1/16+1/16b(n-2)=1/4+1/16+1/64+1/64b(n-3)=。=1/4+(1/4)^2+(1/4)^3+。。。+(1/4)^(n-1)+(1/4)^(n-1)b1=1/4(1+(1/4)^(n-1))/(1-1/4)+(1/4)^(n-1)b1=1/4(1-(1/4)^(n-1))/(1-1/4)+(1/4)^(n-1)b1=1/4-(1/4)^(n-1)/3+2(1/4)^(n-1)/3=1/4+(1/4)^(n-1)/3(2)sn=n/4+(4/3)(1-(1/4)^n)/(1-1/4)=n/4+(16/9)(1-(1/4)^n)=n/4-4^(2-n)/9+16/9sn随n的递增而递增。s1=b1=2/3>0。因此sn≥2/3所以1/sn≤3/2。对于任意k≤3/2有 1/s1≥k。因此m=3/2具体思路大概是这样,