Dãy, chuỗi và hội tụ đều
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| pointwise/ˈpɔɪntwaɪz/ | Converging separately at each x, with N allowed to depend on x. |
| uniform convergence/ˈjuːnɪfɔːm kənˈvɜːdʒəns/ | hội tụ đều |
| conditionally/kənˈdɪʃənəli/ | Convergent, but not absolutely convergent. |
Continuous curves, a jump in the limit
- Each $x^n$ is continuous on $[0,1]$, yet the limit is $0$ for $x < 1$ and $1$ at $x = 1$ — a jump. Which kind of convergence would have prevented it?
- GRE analysis tests sequences, series tests, and the gap between pointwise 逐点 and uniform convergence 一致收敛.
Sequences
- In $\mathbb R$, convergent $\iff$ Cauchy (completeness). In $\mathbb Q$, a Cauchy sequence can head for $\sqrt 2$ and fail to converge.
- Bounded monotone sequences converge. Bounded alone does not, but Bolzano–Weierstrass gives a convergent subsequence: $(-1)^n$ has two subsequential limits.
Series tests
- Absolute convergence implies convergence. $\sum (-1)^n/n$ converges only conditionally 条件地, and rearranging it can change the sum.
- Ratio test: limit $< 1$ converges absolutely, $> 1$ diverges, $= 1$ says nothing ($\sum 1/n$ diverges, $\sum 1/n^2$ converges).
For $\sum n/2^n$, the ratio test gives
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$\frac{n+1}{2n} \to \frac12$.
$\sum (-1)^n/n$ converges, but not absolutely: it converges ____.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Its absolute series is harmonic.
Pointwise versus uniform
- Pointwise: for each $x$, $f_n(x) \to f(x)$; $N$ may depend on $x$. Uniform: $\sup_x |f_n(x) - f(x)| \to 0$; one $N$ for all $x$.
$x^n$ on $[0,1]$: the limit is discontinuous, and $\sup_{[0,1)} x^n = 1$ for every $n$, so convergence is not uniform. On $[0, \tfrac12]$ the supremum error is $(\tfrac12)^n \to 0$: uniform. The domain changed the answer.
![Left: x^n on [0, 1] for growing n, converging only pointwise. Right: the supremum error (1/2)^n on [0, 1/2] falling to 0](/learn-media/gre_mathematics/assets/1301.svg?v=1790958504523250741)
Pointwise only, or uniform?
Decide uniform convergence from the supremum error.
On $[0, \tfrac12]$, what is $\sup |x^3 - 0|$?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
x³ increases on [0, ½], so the supremum of |x³| is at x = ½: (½)³ = 1/8 = 0.125.
Which sequence converges uniformly on $[0, 1]$?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$\sup |x^n/n| = 1/n \to 0$.
What uniform convergence buys
- Uniform limits of continuous functions are continuous; on $[a,b]$, limits and integrals can be swapped.
- Swapping limits and derivatives needs more (uniform convergence of $f_n'$). Weierstrass M-test: $|f_n(x)| \le M_n$ for all $x$ with $\sum M_n < \infty$ gives uniform convergence.
Checking a few fixed $x$ values proves nothing about the supremum: the worst point can move with $n$.
A uniform limit of continuous functions is continuous.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. If continuous fₙ → f uniformly, an ε/3 argument shows f is continuous. Pointwise limits need not be: xⁿ on [0, 1] has a discontinuous limit.
Recap
- Cauchy = convergent in $\mathbb R$; bounded sequences have convergent subsequences.
- Ratio-test limit 1 is inconclusive.
- Uniform means the supremum error tends to 0; it preserves continuity and integrals.
- pointwise: Converging separately at each x, with N allowed to depend on x.
- conditionally: Convergent, but not absolutely convergent.