level 8
记|a_n|=b_n,b_n非负,
则n>N(p)时,b_(n+2)>-b_(n+1)+pb_n,
解特征方程t^2+t-p=0两个根为t_1,t_2,不妨设t_1<0<t_2,由p的取法知t_2>1,
则n>N(p)时,b_(n+2)-t_1*b_(n+1)>t_2*(b_(n+1)-t_1*b_n),
适当增大n使得b_(n+1)-t_1*b_n>0(一定可以取到,不然{b_n}只有有限项非零,极限无意义),
则b_(n+k+1)-t_1*b_(n+k)>(t_2)^k*(b_(n+1)-t_1*b_n),
当k趋于无穷时右端趋于无穷,而左端有界,矛盾
![[扯花]](/static/emoticons/u626fu82b1.png)