Comparison Tests for Convergence · 收敛的比较判别法
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| comparison tests/kəmˈpærɪsn tests/ | 比较判别法 | bǐ jiào pàn bié fǎ |
Judging a series by a known neighbor
- If a mystery series looks like a series you already understand, compare them.
- The comparison tests 比较判别法 measure an unknown positive series against a known one (usually a p-series or geometric series).
- Bigger than something that diverges → diverges. Smaller than something that converges → converges.
- Two flavors: a direct comparison and a limit comparison.
用已知的"邻居"判断级数
- 若一个未知级数看起来像你已经理解的级数,就比较它们。
- 比较判别法把一个未知的正项级数与一个已知的(通常是 p-级数或几何级数)比较。
- 比发散的东西大 → 发散。比收敛的东西小 → 收敛。
- 两种口味:直接比较和极限比较。
Direct Comparison Test
- For positive terms with $0\le a_n\le b_n$:
- If $\sum b_n$ converges, then the smaller $\sum a_n$ converges too.
- If $\sum a_n$ diverges, then the bigger $\sum b_n$ diverges too.
- Squeeze from the right side: bounded above by a convergent series, or below by a divergent one.
直接比较判别法
- 对正项,若 $0\le a_n\le b_n$:
- 若 $\sum b_n$ 收敛,则更小的 $\sum a_n$ 也收敛。
- 若 $\sum a_n$ 发散,则更大的 $\sum b_n$ 也发散。
- 从对的一侧夹:被一个收敛级数从上界住,或被一个发散级数从下界住。
Compare to a known series · 与已知级数比较
A positive series bounded by a known convergent (or divergent) series shares its fate — pick the dominant comparison. · 有界于已知收敛（或发散）级数的正级数与其命运相同 — 选取主导的比较对象。
If $0\le a_n\le b_n$ and $\sum b_n$ converges, then $\sum a_n$... · 若$0\le a_n\le b_n$和$\sum b_n$收敛，则$\sum a_n$...
Smaller than a convergent series → converges. · 小于收敛级数 → 收敛。
The comparison tests require positive terms. · 比较判别法要求各项为正。
They compare magnitudes of positive series. · 它们比较正级数的大小。
Limit Comparison Test
- Sometimes inequalities are awkward; instead compare growth rates.
- Take $\displaystyle L=\lim_{n\to\infty}\dfrac{a_n}{b_n}$. If $L$ is a finite positive number ($0
- Pick $b_n$ from the dominant terms of $a_n$ (e.g. compare $\tfrac{n}{n^3+1}$ with $\tfrac1{n^2}$).
- The limit being positive and finite means they behave alike.
极限比较判别法
- 有时不等式很别扭;那就比较增长率。
- 取 $\displaystyle L=\lim_{n\to\infty}\dfrac{a_n}{b_n}$。若 $L$ 是有限正数($0
- 从 $a_n$ 的主导项挑 $b_n$(如把 $\tfrac{n}{n^3+1}$ 与 $\tfrac1{n^2}$ 比较)。
- 极限为正且有限,意味着它们行为相似。
In the Limit Comparison Test, if $\lim\tfrac{a_n}{b_n}=L$ with $0
A finite positive limit → same convergence behavior. · 有限的正极限 → 相同的收敛行为。
For the Limit Comparison Test to conclude, the limit $L$ must be finite and . · 为使极限比较判别法得出结论，极限$L$必须是有限的且。
$0
Choosing the comparison series
- Look at the dominant behavior of $a_n$ for large $n$ — the highest powers on top and bottom.
- $\dfrac{3n^2+1}{n^4+5}$ behaves like $\dfrac{3n^2}{n^4}=\dfrac{3}{n^2}$ — compare with the p-series $\tfrac1{n^2}$.
- Then either bound it directly or take the limit ratio.
- The right comparison series turns a scary series into a known one.
选择比较级数
- 看 $a_n$ 在大 $n$ 时的主导行为——上下的最高次幂。
- $\dfrac{3n^2+1}{n^4+5}$ 表现得像 $\dfrac{3n^2}{n^4}=\dfrac{3}{n^2}$——与 p-级数 $\tfrac1{n^2}$ 比较。
- 然后要么直接界住,要么取极限之比。
- 正确的比较级数把吓人的级数变成已知的。
To test $\sum\dfrac{3n^2+1}{n^4+5}$, a good comparison series is... · 要测试$\sum\dfrac{3n^2+1}{n^4+5}$，一个好的比较级数是...
Dominant behavior $\tfrac{3n^2}{n^4}=\tfrac{3}{n^2}$ → compare with $\tfrac1{n^2}$. · 主导行为$\tfrac{3n^2}{n^4}=\tfrac{3}{n^2}$ → 与$\tfrac1{n^2}$比较。
Since $\tfrac{1}{n^2+1}<\tfrac1{n^2}$ and $\sum\tfrac1{n^2}$ converges, $\sum\tfrac{1}{n^2+1}$... · 由于$\tfrac{1}{n^2+1}<\tfrac1{n^2}$和$\sum\tfrac1{n^2}$收敛，$\sum\tfrac{1}{n^2+1}$...
Bounded above by a convergent series → converges. · 被收敛级数上界界定 → 收敛。
Comparison tests need positive terms. For Direct Comparison, the inequality must point the right way: to prove convergence, bound above by a convergent series; to prove divergence, bound below by a divergent one. For Limit Comparison, the limit $L$ must be finite and positive ($0
比较判别法需要正项。对直接比较,不等式必须朝对的方向:要证收敛,用收敛级数从上界住;要证发散,用发散级数从下界住。对极限比较,极限 $L$ 必须有限且为正($0
Does $\displaystyle\sum\dfrac{1}{n^2+1}$ converge?
- For large $n$, $\dfrac{1}{n^2+1}$ behaves like $\dfrac{1}{n^2}$ (a convergent p-series).
- Direct: $\dfrac{1}{n^2+1}<\dfrac{1}{n^2}$, and $\sum\tfrac1{n^2}$ converges → so does the smaller series.
- Therefore $\sum\dfrac{1}{n^2+1}$ converges.
$\displaystyle\sum\dfrac{1}{n^2+1}$ 收敛吗?
- 对大 $n$,$\dfrac{1}{n^2+1}$ 表现得像 $\dfrac{1}{n^2}$(一个收敛 p-级数)。
- 直接: $\dfrac{1}{n^2+1}<\dfrac{1}{n^2}$,而 $\sum\tfrac1{n^2}$ 收敛 → 更小的级数也收敛。
- 因此 $\sum\dfrac{1}{n^2+1}$ 收敛。
The comparison tests judge a positive series against a known one. Direct: bound above by a convergent series (→ converges) or below by a divergent one (→ diverges). Limit: if $\lim\frac{a_n}{b_n}$ is finite and positive, both share the same fate. Choose $b_n$ from $a_n$'s dominant terms.
比较判别法用一个已知级数判断正项级数。直接: 用收敛级数从上界住(→ 收敛)或用发散级数从下界住(→ 发散)。极限: 若 $\lim\frac{a_n}{b_n}$ 有限且为正,两者命运相同。从 $a_n$ 的主导项挑 $b_n$。