幂级数的收敛区间.
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/06/26 03:55:30
幂级数的收敛区间.
![](http://img.wesiedu.com/upload/d/3b/d3b0e0045ad5f18470061b38aa7c427a.jpg)
![](http://img.wesiedu.com/upload/d/3b/d3b0e0045ad5f18470061b38aa7c427a.jpg)
![幂级数的收敛区间.](/uploads/image/z/20093348-20-8.jpg?t=%E5%B9%82%E7%BA%A7%E6%95%B0%E7%9A%84%E6%94%B6%E6%95%9B%E5%8C%BA%E9%97%B4.)
(n=1,∞)∑[(2n)!/(n!)² *x^(2n)]
= (n=1,∞)∑[(2n)!/((2n-n)!*(n!)²) *x^(2n)]
= (n=1,∞)∑[C(n,2n) *x^(2n)]
级数通项 u(n) = C(n,2n) *x^(2n)
(n→∞)lim u(n+1)/u(n)
= (n→∞)lim [C(n+1,2n+2)*x^(2n+2)]/[C(n,2n) *x^(2n)]
= (n→∞)lim (2n+2)(2n+1)*x^2/(n+1)^2
= 4x^2
当 4x^2 |x| < 2 时,级数收敛;
因此幂级数的收敛区间为 |x| < 2
= (n=1,∞)∑[(2n)!/((2n-n)!*(n!)²) *x^(2n)]
= (n=1,∞)∑[C(n,2n) *x^(2n)]
级数通项 u(n) = C(n,2n) *x^(2n)
(n→∞)lim u(n+1)/u(n)
= (n→∞)lim [C(n+1,2n+2)*x^(2n+2)]/[C(n,2n) *x^(2n)]
= (n→∞)lim (2n+2)(2n+1)*x^2/(n+1)^2
= 4x^2
当 4x^2 |x| < 2 时,级数收敛;
因此幂级数的收敛区间为 |x| < 2