Alternating Series Test for Convergence · 交错级数判别法判定收敛
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| alternating series/ˈɔːltəneɪtɪŋ ˈsɪəriːz/ | 交错级数 | jiāo cuò jí shù |
Series whose signs flip
- Some series alternate sign: $+,-,+,-,\dots$, like $1-\tfrac12+\tfrac13-\tfrac14+\cdots$.
- These are alternating series 交错级数, and they have their own gentle convergence test.
- The sign flips let terms partly cancel, so alternating series converge more easily.
- One test with two simple conditions handles them.
符号交替翻转的级数
- 有些级数符号交替:$+,-,+,-,\dots$,如 $1-\tfrac12+\tfrac13-\tfrac14+\cdots$。
- 这些是交错级数,它们有自己温和的收敛判别法。
- 符号翻转让各项部分抵消,所以交错级数更容易收敛。
- 一个带两个简单条件的判别法就能处理它们。
The Alternating Series Test
- An alternating series $\sum(-1)^n b_n$ (with $b_n>0$) converges if both:
- 1. the terms decrease: $b_{n+1}\le b_n$, and
- 2. the terms shrink to zero: $\displaystyle\lim_{n\to\infty}b_n=0$.
- If both hold, the alternating series converges. Simple as that.
交错级数判别法
- 交错级数 $\sum(-1)^n b_n$(其中 $b_n>0$)收敛,若两者都成立:
- 1. 各项递减:$b_{n+1}\le b_n$,且
- 2. 各项缩向零:$\displaystyle\lim_{n\to\infty}b_n=0$。
- 若两者都成立,交错级数收敛。就这么简单。
Shrinking alternating terms · 递减的交错项
If the positive part decreases to $0$, the sign flips let the partial sums close in — the series converges. · 若正部递减至$0$，符号翻转使部分和收敛 — 级数收敛。
The Alternating Series Test requires the positive terms $b_n$ to be which? · 交错级数判别法要求正项$b_n$具备哪种性质？
Decreasing and limit $0$. · 递减且极限$0$。
Why the flips help
- Each new term is smaller than the last and opposite in sign, so it partly undoes the previous one.
- The partial sums bounce, but by shrinking amounts — closing in on a limit.
- That's why $1-\tfrac12+\tfrac13-\cdots$ converges, even though $1+\tfrac12+\tfrac13+\cdots$ (harmonic) diverges.
- The alternation is what tames the harmonic terms.
翻转为何有帮助
- 每个新项比上一个小且符号相反,所以它部分抵消了前一个。
- 部分和来回跳,但以缩小的幅度——向一个极限靠拢。
- 这就是为何 $1-\tfrac12+\tfrac13-\cdots$ 收敛,尽管 $1+\tfrac12+\tfrac13+\cdots$(调和)发散。
- 是交替驯服了调和的各项。
The alternating harmonic series $\sum\tfrac{(-1)^{n+1}}{n}$... · 交错调和级数$\sum\tfrac{(-1)^{n+1}}{n}$...
$b_n=\tfrac1n$ decreases to $0$ → converges. · $b_n=\tfrac1n$递减至$0$ → 收敛。
The sign ____ let the partial sums close in on a limit. · 符号 ____ 使部分和收敛于某个极限。
Alternation causes partial cancellation. · 交错导致部分抵消。
$\sum\tfrac{(-1)^n}{n}$ converges even though $\sum\tfrac1n$ diverges. · $\sum\tfrac{(-1)^n}{n}$ 收敛，尽管 $\sum\tfrac1n$ 发散。
The alternation tames the harmonic terms. · 交错驯服了调和项。
Applying it cleanly
- Strip off the $(-1)^n$ and check the positive part $b_n$ for the two conditions.
- Is $b_n$ decreasing? Does $b_n\to0$? If yes to both, converges.
- If $b_n$ does not go to $0$, the series diverges (by the nth Term Test instead).
- The test is quick once you isolate $b_n$.
干净地应用它
- 剥掉 $(-1)^n$,对正部 $b_n$ 检查两个条件。
- $b_n$ 递减吗?$b_n\to0$ 吗?若两者都是,收敛。
- 若 $b_n$ 不趋于 $0$,级数发散(改用项判别法)。
- 一旦孤立出 $b_n$,判别法就很快。
You check the two conditions on the positive part $b_n$ after removing $(-1)^n$. · 检查正部上的两个条件 $b_n$ 移除后 $(-1)^n$.
Isolate $b_n$ first. · 先分离出 $b_n$。
If $b_n\not\to0$ for an alternating series, it... · 如果是交错级数且 $b_n\not\to0$ ...
Terms not $\to0$ → diverges regardless of alternation. · 通项不趋于 $\to0$ → 无论是否交错均发散。
Apply the two conditions to the positive terms $b_n$ (after removing the $(-1)^n$). Both are needed: decreasing and limit $0$. If $b_n\not\to0$, the series diverges (nth Term Test). And this test proves convergence of the alternating series; it says nothing about $\sum b_n$ or $\sum|a_n|$ (see absolute vs. conditional, lesson 10.9).
对正项 $b_n$(去掉 $(-1)^n$ 后)应用两个条件。两者都需要:递减且极限为 $0$。若 $b_n\not\to0$,级数发散(项判别法)。而此判别法证明的是交错级数的收敛;它对 $\sum b_n$ 或 $\sum|a_n|$ 什么都不说(见绝对与条件,10.9 课)。
Does the alternating harmonic series $1-\tfrac12+\tfrac13-\tfrac14+\cdots$ converge?
- Positive part $b_n=\tfrac1n$. Decreasing? Yes, $\tfrac1{n+1}<\tfrac1n$. Limit $0$? Yes, $\tfrac1n\to0$.
- Both conditions hold, so the series converges by the Alternating Series Test.
- (Even though the plain harmonic series diverges.)
交错调和级数 $1-\tfrac12+\tfrac13-\tfrac14+\cdots$ 收敛吗?
- 正部 $b_n=\tfrac1n$。递减? 是,$\tfrac1{n+1}<\tfrac1n$。极限 $0$? 是,$\tfrac1n\to0$。
- 两个条件都成立,所以由交错级数判别法级数收敛。
- (尽管普通调和级数发散。)
The Alternating Series Test: $\sum(-1)^n b_n$ (with $b_n>0$) converges if the terms $b_n$ are decreasing and $b_n\to0$. Check both on the positive part $b_n$. The alternation lets series like $\sum\frac{(-1)^n}{n}$ converge where the non-alternating version diverges.
交错级数判别法:$\sum(-1)^n b_n$(其中 $b_n>0$)收敛,若各项 $b_n$ 递减且 $b_n\to0$。对正部 $b_n$ 检查两者。交替让像 $\sum\frac{(-1)^n}{n}$ 这样的级数收敛,而非交错版本却发散。