公式及其必然成立的原因
柯西–阿達馬定理說，sum c_n (x - a)^n 的半徑 R 由 1/R = limsup of |c_n|^(1/n)（當 n -> 無窮）給出，約定 1/0 = 無窮、1/無窮 = 0。原因是：在某點 x 處應用根值判別法恰好讀出這個量。
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).
比值判別法，更簡單的特例
實踐中比值判別法往往更簡單。當極限 lim |c_{n+1} / c_n| 存在時，它等於 limsup |c_n|^(1/n)，於是可用 R = lim |c_n / c_{n+1}|。比值判別法不如根值判別法一般——即使柯西–阿達馬有效，它也可能給不出極限——但對階乘和乾淨的閉式係數，它是快速途徑。
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.端點決定區間
柯西–阿達馬給出 R，但對 |x - a| = R 處一無所知。完整的收斂區間要靠分別檢驗每個端點求得。R = 1 的級數 sum x^n / n 是兩端可以不一致的經典例證。
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.