T.4 · Sequências, séries e convergência uniforme
GRE Subject Test · GRE Mathematics Tópico 13 2:52 Narração em inglês · Legendas em inglês + 中文 gravadas
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This unit uses Cauchy, monotone convergence and subsequence criteria; distinguishes pointwise from uniform convergence of functions; and checks hypotheses before interchanging limits and integrals.
本单元使用柯西准则、单调收敛准则和子列准则；区分函数的逐点收敛与一致收敛；并在交换极限与积分之前检查条件。
Every real Cauchy sequence converges because R is complete; in Q one can approach an irrational and fail to converge.
因为 R 完备，每个实柯西列都收敛；在 Q 中，柯西列可能趋向无理数而不收敛。
A bounded monotone sequence converges.
有界单调数列收敛。
A bounded sequence need not converge, but Bolzano-Weierstrass gives a convergent subsequence.
有界数列不一定收敛，但波尔查诺–魏尔斯特拉斯定理保证有收敛子列。
Absolute convergence implies convergence; a conditionally convergent series cannot be rearranged freely.
绝对收敛蕴含收敛；条件收敛的级数不能随意重排。
The alternating harmonic series converges, but its absolute series diverges.
交错调和级数收敛，但其绝对值级数发散。
A ratio limit of exactly 1 is inconclusive: the sum of 1 over n and the sum of 1 over n squared both have it.
比值极限恰好为 1 时无法判断：∑1/n 与 ∑1/n² 的比值极限都是 1。
Pointwise convergence chooses N separately for each x; uniform convergence chooses one N for every x, so check the supremum of the error.
逐点收敛对每个 x 分别选 N；一致收敛对所有 x 选同一个 N，所以要检查误差的上确界。
A uniform limit of continuous functions is continuous, and on a closed interval limit and integral can be exchanged.
连续函数的一致极限是连续的，在闭区间上可以交换极限与积分。
Derivatives need more.
导数则需要更多条件。
On 0, 1, f n of x equals x to the n tends to 0 for x below 1 and to 1 at x equals 1.
在 上，fₙ(x) = xⁿ 在 x < 1 时趋于 0，在 x = 1 时趋于 1。
The limit is discontinuous, so convergence cannot be uniform: the supremum error stays 1.
极限不连续，所以不可能一致收敛：上确界误差一直是 1。
On 0, a with a below 1, the supremum is a to the n, which tends to zero: uniform.
在 （a < 1）上，上确界是 aⁿ，趋于零：一致收敛。
Checking several fixed x values proves neither a supremum bound nor uniform convergence.
检查几个固定的 x 值，既不能证明上确界的界，也不能证明一致收敛。
The point producing the largest error may move with n.
误差最大的点可能随 n 移动。
On 0, 1 the limit is discontinuous, so continuous f n cannot converge uniformly.
在 上极限不连续，所以连续的 fₙ 不可能一致收敛。
On 0, a the supremum is a to the n, which tends to 0: uniform.
在 上上确界是 aⁿ，趋于 0：一致收敛。
The p-series with p equals 2 converges absolutely; the alternating harmonic series converges only conditionally.
p = 2 的 p 级数绝对收敛；交错调和级数只是条件收敛。
The supremum of g n is 1 over n, so the functions converge uniformly to zero.
gₙ 的上确界是 1/n，所以函数一致收敛到零。
But the derivative is cos n x, which equals 1 at x equals 0 for every n, while the limit's derivative is 0.
但导数是 cos nx，在 x = 0 处对每个 n 都等于 1，而极限的导数是 0。
Uniform convergence alone does not justify differentiating a limit.
仅有一致收敛，不能对极限求导。
Check yourself.
自我检测。
x to the n is not uniform on 0, 1, but is uniform on 0, a half.
xⁿ 在 上不一致收敛，但在 上一致收敛。
A ratio limit of 1 is inconclusive.
比值极限为 1 时无法判断。
The alternating harmonic series converges conditionally.
交错调和级数条件收敛。
And the derivatives of sin n x over n do not converge to zero.
sin(nx)/n 的导数不收敛到零。