lim(n→∞) [(n^3-1）/(3*n^2+n)-an-b]
=lim(n→∞) [(1-3a)n^3-(a+3b)n^2-bn-1]/(3*n^2+n)]=0
所以
1-3a=0,且a+3b=0
从而
a=1/3,b=-1/9.