※ 引述《tasukuchiyan (Tasuku)》之銘言：
: Investigate the behavior of Σan if
: an = 1/(1+z^n)，for complex values of z
[Rudin, Principles of Mathematical Analysis-exercise. 6, Chap. 3.]
Let a_n ＝ 1/(1＋ z^n), and we consider three cases as follows.
(i) |z| ＜ 1 (ii) |z| ＞ 1, and (iii) |z| ＝ 1.
In case (i), lim a_n ＝ 1. So, we know that Σ a_n diverges by Theorem 3.23.
(Sometimes, we call Theorem 3.23, n-test.)
In case (ii), write z ＝ r e^(iθ), where r ＞ 1. Then
| 1 ＋ z^n ｜ ＝ | 1 ＋ r^n * e^(inθ) | ≧ r^n.
Hence Σ a_n converges (converges absolutely) by the comparison test on the
convergene of Σ 1/r^n, where r ＞ 1.
In case (iii), It is clear that as z ＝ 1, Σ a_n diverges and as z ＝ －1,
it does not make sense. As a result, it remains to consider z ≠1 and ≠-1
as follows. Write z ＝ cosθ ＋ i sinθ and thus, we have
1 ＋ z^n ＝ (1＋cos nθ)＋i(sin nθ).
It follows that
1 1
|a_n|^2 ＝ |-------------------------|^2 ＝ ----------------------
(1＋cos nθ)＋i(sin nθ) 2 ＋ 2 cos nθ
≧ 1/4.
It means that {a_n} cannot converge to zero. That is, Σ a_n diverges in this
case.
NOTE. 在 case (iii) 裡，當 z ＝ i, z^2 ＝ －1. 這時候分母也會等於 0. 假使
我們接受 |1/0| ＝ ∞, 那麼
|a_n|^2 ≧ 1/4
still makes sense.
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※ 編輯: math1209 來自: 220.133.4.14 (12/03 07:32)