Question

已知幂级数2n-1/2的n次方

求幂级数[∞∑n=1] (2n-1/2^n)*X^(2n-2)的收敛区间,还有当求出收敛半径的时候,如何
快速判断收敛半径的端点是否使得该幂级数收敛啊?

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∵lim(n->∞)│[(2n+1)x^(2n)/2^(n+1)]/[(2n-1)x^(2n-2)/2^n]│
=lim(n->∞){[(2n+1)x²]/[2(2n-1)]}
=lim(n->∞){[(2+1/n)x²]/[2(1-1/n)]}
=x²/2
令x²/2│x│