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【趣味数学题】比值审敛法

2021-10-13 19:40 作者:AoiSTZ23  | 我要投稿

郑涛(Tao Steven Zheng)著

【问题】

题一:考虑下面的无穷级数。使用比值审敛法(ratio test)来确定各无穷级数是收敛还是发散的级数。

(1)%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bn%7D%7B2%5En%7D%20

(2)%20%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B2%5En%7D%7Bn%7D%20

题二: 求以下无穷级数的收敛半径(radius of convergence)和收敛区间(interval of convergence)。


%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B4%5En%7D%7Bn%7D(x-3)%5En%20

【题解】

题一
(1)
%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Ba_n%7D%20%5Cright%7C%20

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B%5Cfrac%7Bn%2B1%7D%7B2%5E%7Bn%2B1%7D%7D%7D%7B%5Cfrac%7Bn%7D%7B2%5En%7D%7D%5Cright%7C

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7Bn%2B1%7D%7B2n%7D%20%5Cright%7C%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20

由于该极限小于1,因此级数收敛。

(2)
%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Ba_n%7D%20%5Cright%7C%20

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B%5Cfrac%7B2%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%7D%7B%5Cfrac%7B2%5En%7D%7Bn%7D%7D%20%5Cright%7C%20

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B2n%7D%7Bn%2B1%7D%20%5Cright%7C%20%3D%202%20

由于该极限大于1,因此级数发散。

题二

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Ba_n%7D%20%5Cright%7C%20%3C%201%20

%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B4%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%20%7B(x-3)%7D%5E%7Bn%2B1%7D%20%5Ccdot%20%5Cfrac%7Bn%7D%7B4%5En%20%7B(x-3)%7D%5E%7Bn%7D%7D%20%5Cright%7C%20%3C%201

%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B4n%7D%7Bn%2B1%7D(x-3)%20%5Cright%7C%20%3C%201

%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft%7C%20%5Cfrac%7B4n%7D%7Bn%2B1%7D%5Cright%7C%20%7Cx-3%7C%20%3C%201

%204%7Cx-3%7C%20%3C%201%20

%20%7Cx-3%7C%20%3C%20%5Cfrac%7B1%7D%7B4%7D

我们发现收敛半径为 %20R%20%3D%20%5Cfrac%7B1%7D%7B4%7D

求解收敛区间


%5Cfrac%7B-1%7D%7B4%7D%20%3C%20x-3%20%3C%20%5Cfrac%7B1%7D%7B4%7D

%5Cfrac%7B11%7D%7B4%7D%20%3C%20x%20%3C%20%5Cfrac%7B13%7D%7B4%7D%20


是比值审敛法得出的结论,但是我们还要确定每个端点是否收敛。所以我们必须应用其他方法。

端点(一): x%3D%5Cfrac%7B11%7D%7B4%7D

%20%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B4%5En%7D%7Bn%7D%7B%5Cleft(%5Cfrac%7B-1%7D%7B4%7D%5Cright)%7D%5E%7Bn%7D

%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B(-1)%5En%7D%7Bn%7D%20

通过交错级数判别法(alternating series test),该级数收敛。


端点(二):  x%20%3D%20%5Cfrac%7B13%7D%7B4%7D

%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B4%5En%7D%7Bn%7D%7B%5Cleft(%5Cfrac%7B1%7D%7B4%7D%5Cright)%7D%5E%7Bn%7D

%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7Bn%7D

这是调和级数(harmonic series),然后这里证明该级数发散。因此,

%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B4%5En%7D%7Bn%7D(x-3)%5En

收敛半径为 %5Cfrac%7B11%7D%7B4%7D%20%5Cle%20x%20%3C%20%5Cfrac%7B13%7D%7B4%7D.




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