由题可知，f(0)=f'(0)=0,然后将f(x)在x=0处展开成带有peano余项的公式，即f(x)=f(0)+f'(0)x+f"(0)/2*x^2+O(x^2)=f''(0)/2*x^2+O(x^2),所以f(1/n)=f''(0)/2*1/n^2+O(1/n^2),所以原题就等价于求1/n^2*x^n的收敛域，所以收敛域应该是[-1，1 ].[draw]a11i21321421321221322521323b21323f2132401e2431724316243a11i1d22b21422721c227a11i22721b22621b22022921f23721f23ba11i22023321f23321f23821f23c21f23f21f240223240227241a11i22e225 ...