有界与单调
有界数列是其各项永不逃出固定窗口的数列:存在数 M 使对所有 n 有 |a_n| <= M。第一个基本事实:每个收敛数列都有界。(越过某个 N,各项落在 L 的 1 范围内,故被 |L| + 1 控制;前面有限多项有一个最大绝对值;取最大者即可。)逆命题不成立——(-1)^n 有界却发散——所以仅有界并不够。
单调数列只朝一个方向走:递增(对所有 n 有 a_{n+1} >= a_n)或递减(对所有 n 有 a_{n+1} <= a_n)。单调数列无法振荡,故它们唯一可能不收敛的方式,就是奔向无穷。这一观察是下一个定理的核心。
单调收敛定理
单调收敛定理说:递增且有上界的数列收敛,其极限是各项的上确界。(对称地,递减且有下界的数列收敛于下确界。)这是第一个无需有人事先告诉你答案就召唤出极限的定理——它之所以成立,恰因实数借助最小上界性质而完备。
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极限的运算与夹逼
一旦你知道若干极限,极限的运算便让你组合它们而无需回到 epsilon。若 a_n -> A 且 b_n -> B,则:a_n + b_n -> A + B;a_n - b_n -> A - B;a_n b_n -> AB;且当 B 非 0(且各项最终非零)时 a_n / b_n -> A/B。每条用 epsilon 证一次,便可永远重用——这就是认真建立定义的回报。
夹逼定理是另一员主力:若最终有 a_n <= b_n <= c_n,且 a_n -> L、c_n -> L,则 b_n -> L 也成立。它通过把未知数列夹在两个已知数列之间,来求那些无法直接攻克的极限。
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.