有界與單調
有界數列是其各項永不逃出固定窗口的數列:存在數 M 使對所有 n 有 |a_n| <= M。第一個基本事實:每個收斂數列都有界。(越過某個 N,各項落在 L 的 1 範圍內,故被 |L| + 1 控制;前面有限多項有一個最大絕對值;取最大者即可。)逆命題不成立——(-1)^n 有界卻發散——所以僅有界並不夠。
單調數列只朝一個方向走:遞增(對所有 n 有 a_{n+1} >= a_n)或遞減(對所有 n 有 a_{n+1} <= a_n)。單調數列無法振盪,故它們唯一可能不收斂的方式,就是奔向無窮。這一觀察是下一個定理的核心。
單調收斂定理
單調收斂定理說:遞增且有上界的數列收斂,其極限是各項的上確界。(對稱地,遞減且有下界的數列收斂於下確界。)這是第一個無需有人事先告訴你答案就召喚出極限的定理——它之所以成立,恰因實數藉助最小上界性質而完備。
Theorem: (a_n) increasing and bounded above => converges to L = sup{a_n}.
Proof:
The set S = {a_n : n in N} is nonempty and bounded above,
so by the least upper bound property L = sup S exists.
Let e > 0. Since L - e is NOT an upper bound (L is the LEAST one),
some term a_N satisfies a_N > L - e.
For all n > N, monotonicity gives a_n >= a_N > L - e.
Also a_n <= L for every n (L is an upper bound), so a_n <= L < L + e.
Combining: L - e < a_n <= L, hence |a_n - L| < e.
Therefore a_n -> L. QED極限的運算與夾逼
一旦你知道若干極限,極限的運算便讓你組合它們而無需回到 epsilon。若 a_n -> A 且 b_n -> B,則:a_n + b_n -> A + B;a_n - b_n -> A - B;a_n b_n -> AB;且當 B 非 0(且各項最終非零)時 a_n / b_n -> A/B。每條用 epsilon 證一次,便可永遠重用——這就是認真建立定義的回報。
夾逼定理是另一員主力:若最終有 a_n <= b_n <= c_n,且 a_n -> L、c_n -> L,則 b_n -> L 也成立。它通過把未知數列夾在兩個已知數列之間,來求那些無法直接攻克的極限。
Evaluate d_n = (3n^2 + 2n) / (n^2 + 5) using the algebra of limits. Divide top and bottom by n^2: d_n = (3 + 2/n) / (1 + 5/n^2). Known facts: 2/n -> 0 and 5/n^2 -> 0 (from 1/n -> 0 and products). Numerator -> 3 + 0 = 3. Denominator -> 1 + 0 = 1 (nonzero!). By the quotient rule: d_n -> 3/1 = 3. Squeeze example: e_n = (sin n) / n. Since -1 <= sin n <= 1, we have -1/n <= e_n <= 1/n. Both -1/n -> 0 and 1/n -> 0, so by the squeeze e_n -> 0.