q=a(n+1)/a(n)=[(n+2)*0.9^(n+1)]/[(n+1)*0.9^n]=
=9(n+2)/10(n+1),
当n<8时,9(n+2)/10(n+1)>1,a(n+1)>a(n);
当n=8时,9(n+2)/10(n+1)=1,a(8)=a(9);
当n>8时,9(n+2)/10(n+1)<1,a(n+1)取N=8或9,就有对于任意自然数n都有an<=aN成立.