令t=x+4, 则x=t-4, 将f(x)展开成t 幂级数即可。 f(x)=1/(x+1)(x+2)=1/(x+1)-1/(x+2) =1/(t-3)-1/(t-2) =-1/(1-t/3)+1/(1-t/2) =-[1+t/3+t2/32+...]+[1+t/2+t2/22+.....], 收敛域为|t|<2 =t/6+(1/22-1/32)t2+(1/23-1/33)t3+.....
