- 函数的问题将函数f(x)=1/x^2+x
- 将f(x)=1/x^2+x-6展开成x的幂级数，并写出收敛区间
- 解：f(x)=1/x^2+x-6=1/(x+3)(x-2)=(1/5)[1/(x-2)-1/(x+3)]
=(1/5){(-1/2)/[1-(x/2)]-(1/3)/[1-(-x/3)]}
=(1/5)[(-1/2)∑(x/2)^n-(1/3)∑(-x/3)^n]
=(1/5)[(-1/2)(1/2)^n-(1/3)(-1/3)^n]∑<0,∞>(x^n)
=(1/5)[-(1/2)^(n+1)+(-1/3)^(n+1)]∑<0,∞>(x^n)
-1＜x/2＜1，-1＜-x/3＜1，即-2＜x＜2
故收敛区间为(-2,2)。