柯西–阿达马：计算收敛半径

公式及其必然成立的原因

柯西–阿达马定理说，sum c_n (x - a)^n 的半径 R 由 1/R = limsup of |c_n|^(1/n)（当 n -> 无穷）给出，约定 1/0 = 无穷、1/无穷 = 0。原因是：在某点 x 处应用根值判别法恰好读出这个量。

Apply the root test to the absolute series sum |c_n| |x - a|^n.

Let L = limsup |c_n|^(1/n).  The n-th root of the n-th term is

   ( |c_n| |x - a|^n )^(1/n) = |c_n|^(1/n) * |x - a|.

Take limsup over n  ->  L * |x - a|.

Root test:
   if L * |x - a| < 1  the series converges absolutely;
   if L * |x - a| > 1  the n-th term does not -> 0, so it diverges.

The threshold is L * |x - a| = 1, i.e. |x - a| = 1/L.
That threshold is exactly the radius:  R = 1/L = 1 / limsup |c_n|^(1/n).

根值判别法把半径化为一行计算。

比值判别法，更简单的特例

实践中比值判别法往往更简单。当极限 lim |c_{n+1} / c_n| 存在时，它等于 limsup |c_n|^(1/n)，于是可用 R = lim |c_n / c_{n+1}|。比值判别法不如根值判别法一般——即使柯西–阿达马有效，它也可能给不出极限——但对阶乘和干净的闭式系数，它是快速途径。

Example: R for sum x^n / n!  (coefficients c_n = 1/n!).

Ratio test on |c_{n+1} / c_n|:
   |c_{n+1} / c_n| = (1/(n+1)!) / (1/n!) = n! / (n+1)! = 1/(n+1) -> 0.

So 1/R = 0, hence R = infinity.  Converges for all x.    (the exp series)

Example: R for sum n! x^n  (c_n = n!).
   |c_{n+1} / c_n| = (n+1)!/n! = n+1 -> infinity.
So 1/R = infinity, hence R = 0.  Converges only at x = 0.

Example: R for sum x^n / n  (c_n = 1/n).
   |c_{n+1}/c_n| = n/(n+1) -> 1.  So R = 1.

由相邻系数之比得出三个半径。

端点决定区间

柯西–阿达马给出 R，但对 |x - a| = R 处一无所知。完整的收敛区间要靠分别检验每个端点求得。R = 1 的级数 sum x^n / n 是两端可以不一致的经典例证。

Series sum_{n>=1} x^n / n,  radius R = 1, center 0.

Right endpoint x = 1:  sum 1/n = harmonic series -> DIVERGES.

Left endpoint x = -1:  sum (-1)^n / n.
   This is alternating; terms 1/n decrease monotonically to 0.
   By the alternating series test -> CONVERGES (conditionally).

Conclusion: interval of convergence is [-1, 1),
   closed on the left, open on the right.  The two ends differ.

半径相同，端点行为相反——务必检验两端。