f(x)=(x^2)/(3^x),x属于正整数
f(1)=1/3,f(2)=4/9,
下面证明当x≥2,x∈N*时,
f(x+1)0恒成立
f(x+1)/f(x)
=[(x+1)^2/3^(x+1)]/(x^2/3^x)
=(x+1)^2/(3x^2)
=(x^2+2x+1)/(3x^2)
(x^2+2x+1)/(3x^2)-1
=（-2x^2+2x+1)/(3x^2)
=[-2(x-1/2)^2+3/2]/(3x^2)
分子g(x)=-2(x-1/2)^2+3/2
在(1/2,+∞)为减函数
∵x≥2
∴g(x)≤g(2)=-3