Integral Test for Convergence · 积分判别法判定收敛
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Integral Test/ˈɪntɪɡrəl test/ | 积分判别法 | jī fēn pàn bié fǎ |
A series and an integral that rise or fall together
- If a series' terms come from a nice decreasing function, its fate matches that function's integral.
- The Integral Test 积分判别法 compares $\sum a_n$ with $\int f(x)\,dx$, where $a_n=f(n)$.
- Picture the terms as rectangle areas — they're bounded by the area under the curve.
- So the improper integral converging (or not) forces the series to do the same.
级数与积分一同升降
- 若级数的各项来自一个漂亮的递减函数,它的命运与那个函数的积分相同。
- 积分判别法把 $\sum a_n$ 与 $\int f(x)\,dx$ 比较,其中 $a_n=f(n)$。
- 把各项想成矩形面积——它们被曲线下的面积所界定。
- 所以反常积分收敛(或不收敛)迫使级数做同样的事。
The conditions
- The test needs $f$ to be positive, continuous, and decreasing for $x\ge$ some point.
- Then define $a_n=f(n)$ and compare with the improper integral.
- These conditions let the rectangles sandwich the area under $f$.
- Check all three before applying — a non-decreasing $f$ breaks the comparison.
条件
- 判别法需要 $f$ 在 $x\ge$ 某点起正、连续、递减。
- 然后定义 $a_n=f(n)$ 并与反常积分比较。
- 这些条件让矩形夹住 $f$ 下的面积。
- 套用前检查全部三个——非递减的 $f$ 会破坏比较。
The Integral Test requires $f$ to be which on $[1,\infty)$? · 积分判别法要求$f$在$[1,\infty)$上是哪种性质？
Positive, continuous, decreasing — not necessarily a polynomial. · 正、连续、递减 — 不一定是多项式。
The Integral Test connects a series to an integral by setting $a_n=f(\ \_\_\ )$. · 积分判别法通过设定$a_n=f(\ \_\_\ )$将级数与积分联系起来。
The terms are the function sampled at integers. · 各项是该函数在整数处的采样值。
The conclusion
- $\displaystyle\sum a_n$ and $\displaystyle\int_1^\infty f(x)\,dx$ both converge or both diverge — same fate.
- Compute the improper integral: if it's finite, the series converges; if infinite, it diverges.
- (The series' sum is not the integral's value — they only share convergence, not the number.)
- One improper integral settles the whole series.
结论
- $\displaystyle\sum a_n$ 与 $\displaystyle\int_1^\infty f(x)\,dx$ 同时收敛或同时发散——命运相同。
- 计算反常积分:若有限,级数收敛;若无穷,发散。
- (级数的和不是积分的值——它们只共享收敛性,而非那个数。)
- 一个反常积分就定下整个级数。
The Integral Test says $\sum a_n$ and $\int_1^\infty f\,dx$... · 积分判别法指出$\sum a_n$和$\int_1^\infty f\,dx$...
They share fate, not value. · 它们命运相同，而非数值相同。
The value of the improper integral equals the sum of the series. · 反常积分的值等于级数的和。
They share convergence, not the value ($\sum\tfrac1{n^2}=\tfrac{\pi^2}{6}\neq1$). · 它们共享收敛性，而非数值 ($\sum\tfrac1{n^2}=\tfrac{\pi^2}{6}\neq1$)。
Why it works
- The rectangles of area $a_n$ fit just above or below the curve $y=f(x)$.
- So the series is squeezed between $\int f$ and $\int f + a_1$ — finite integral means finite series.
- This is a rigorous "area comparison," turning a sum into an integral you can evaluate.
- It's especially handy for terms like $\tfrac1{n^p}$ and $\tfrac1{n\ln n}$.
为何有效
- 面积为 $a_n$ 的矩形正好嵌在曲线 $y=f(x)$ 的上方或下方。
- 所以级数被夹在 $\int f$ 与 $\int f + a_1$ 之间——积分有限意味着级数有限。
- 这是严格的"面积比较",把一个和变成你能求值的积分。
- 它对像 $\tfrac1{n^p}$ 和 $\tfrac1{n\ln n}$ 这样的项尤其好用。
Rectangles bounded by the curve · 由曲线界定的矩形
y = 1/x²
The term-rectangles fit against the area under $f$, so the series and the integral converge or diverge together. · 项-矩形贴合在$f$下方的面积旁，因此级数和积分同时收敛或发散。
Since $\int_1^\infty \tfrac1{x^2}\,dx=1$ (finite), the series $\sum\tfrac1{n^2}$... · 由于$\int_1^\infty \tfrac1{x^2}\,dx=1$ (有限)，级数$\sum\tfrac1{n^2}$...
Finite integral → convergent series. · 有限积分 → 收敛级数。
If $\int_1^\infty f\,dx$ diverges (to $\infty$), then $\sum a_n$... · 若$\int_1^\infty f\,dx$发散 (至$\infty$)，则$\sum a_n$...
Same fate: divergent integral → divergent series. · 命运相同：发散积分 → 发散级数。
The Integral Test needs $f$ positive, continuous, and decreasing — verify all three. And the test shares only convergence/divergence, not the value: $\sum a_n \neq \int_1^\infty f\,dx$ in general. Don't report the integral's value as the series' sum.
积分判别法需要 $f$ 正、连续、递减——三个都要验证。而判别法只共享收敛/发散,而非值:一般 $\sum a_n \neq \int_1^\infty f\,dx$。别把积分的值报告为级数的和。
Test $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}$ with the Integral Test.
- $f(x)=\tfrac1{x^2}$ is positive, continuous, decreasing for $x\ge1$. ✓
- $\displaystyle\int_1^\infty \tfrac1{x^2}\,dx=1$ (finite, from lesson 6.13).
- The integral converges, so the series converges. (Its sum is $\tfrac{\pi^2}{6}$, not $1$.)
用积分判别法检验 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}$。
- $f(x)=\tfrac1{x^2}$ 在 $x\ge1$ 正、连续、递减。✓
- $\displaystyle\int_1^\infty \tfrac1{x^2}\,dx=1$(有限,见 6.13 课)。
- 积分收敛,所以级数收敛。(它的和是 $\tfrac{\pi^2}{6}$,不是 $1$。)
The Integral Test: if $f$ is positive, continuous, and decreasing with $a_n=f(n)$, then $\sum a_n$ and $\int_1^\infty f(x)\,dx$ share the same fate (both converge or both diverge). Evaluate the improper integral to decide — but its value is not the series' sum.
积分判别法:若 $f$ 正、连续、递减且 $a_n=f(n)$,则 $\sum a_n$ 与 $\int_1^\infty f(x)\,dx$ 命运相同(同时收敛或同时发散)。求反常积分来判定——但它的值不是级数的和。