Defining Convergent and Divergent Infinite Series · 定义收敛与发散的无穷级数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| infinite series/ˈɪnfɪnət ˈsɪəriːz/ | 无穷级数 | wú qióng jí shù |
| converges/kənˈvɜːdʒɪz/ | 收敛 | shōu liǎn |
| diverges/daɪˈvɜːdʒɪz/ | 发散 | fā sàn |
| partial sum/ˈpɑːʃl sʌm/ | 部分和 | bù fèn hé |
Adding up infinitely many numbers
- Can you add up infinitely many terms and get a finite total? Sometimes yes.
- An infinite series 无穷级数 is the sum $a_1+a_2+a_3+\cdots$ of the terms of a sequence.
- It converges 收敛 if the running total approaches a finite number, and diverges 发散 otherwise.
- This unit is all about deciding which — and, when it converges, to what.
把无穷多个数加起来
- 你能把无穷多个项加起来得到一个有限的总和吗?有时能。
- 无穷级数是一个数列各项的和 $a_1+a_2+a_3+\cdots$。
- 若累计总和趋近一个有限数,它就收敛,否则发散。
- 这一单元就是判定收敛与否——以及收敛时收敛到什么。
Partial sums are the key
- Add the terms one at a time: the partial sum 部分和 $S_n=a_1+a_2+\cdots+a_n$ is the total of the first $n$ terms.
- These partial sums form their own sequence $S_1,S_2,S_3,\dots$.
- The series' value is defined as the limit of the partial sums: $\displaystyle\sum a_n=\lim_{n\to\infty}S_n$.
- So an infinite sum is really a limit — the same limit idea from Unit 1.
部分和是关键
- 一项一项地加:部分和 $S_n=a_1+a_2+\cdots+a_n$ 是前 $n$ 项的总和。
- 这些部分和构成它们自己的数列 $S_1,S_2,S_3,\dots$。
- 级数的值定义为部分和的极限:$\displaystyle\sum a_n=\lim_{n\to\infty}S_n$。
- 所以无穷和其实是一个极限——与第 1 单元同样的极限想法。
Terms of a converging series · 收敛级数的项
The terms of $1+\tfrac12+\tfrac14+\cdots$ shrink geometrically, so the partial sums settle at a finite limit. · $1+\tfrac12+\tfrac14+\cdots$的项按几何规律递减，因此部分和收敛于有限极限。
The value of an infinite series is defined as the limit of its... · 无穷级数的值定义为其...的极限。
$\sum a_n=\lim_{n\to\infty}S_n$.
The sum of the first $n$ terms, $S_n$, is called the $n$th ____ sum. · 前$n$项之和$S_n$被称为第$n$个____和。
The series is the limit of these partial sums. · 该级数是这些部分和的极限。
Converge vs. diverge
- Converges: the partial sums $S_n$ approach a finite limit $L$ — that $L$ is the series' sum.
- Diverges: the partial sums grow without bound, or bounce and never settle.
- $1+\tfrac12+\tfrac14+\tfrac18+\cdots$ converges (to $2$); $1+1+1+\cdots$ diverges.
- Convergence is about the partial sums settling down, not the terms.
收敛 vs 发散
- 收敛: 部分和 $S_n$ 趋近一个有限极限 $L$——那个 $L$ 就是级数的和。
- 发散: 部分和无界增长,或来回跳、永不安定。
- $1+\tfrac12+\tfrac14+\tfrac18+\cdots$ 收敛(到 $2$);$1+1+1+\cdots$ 发散。
- 收敛说的是部分和安定下来,而非各项。
The partial sums of $1+\tfrac12+\tfrac14+\cdots$ climb toward what limit? · $1+\tfrac12+\tfrac14+\cdots$的部分和趋向于哪个极限？
A geometric series with ratio $\tfrac12$ sums to $2$. · 公比为$\tfrac12$的几何级数之和为$2$。
Which series clearly diverges? · 哪个级数明显发散？
Terms don't shrink to $0$, so it diverges. · 项不趋近于$0$，故发散。
A series converges when its partial sums... · 当部分和...时，级数收敛。
Convergence = partial sums approach a finite $L$. · 收敛 = 部分和趋近于有限$L$。
Terms shrinking is necessary, not enough
- For a series to converge, its terms must shrink to $0$ — otherwise the total can't settle.
- But shrinking terms alone don't guarantee convergence (the harmonic series $\sum\tfrac1n$ has terms $\to0$ yet diverges).
- So "$a_n\to0$" is a necessary condition, not a sufficient one.
- The tests in this unit are the tools for the harder cases.
各项趋零是必要,不是充分
- 级数要收敛,它的各项必须缩向 $0$——否则总和无法安定。
- 但各项缩小本身不保证收敛(调和级数 $\sum\tfrac1n$ 各项 $\to0$ 却发散)。
- 所以"$a_n\to0$"是必要条件,不是充分条件。
- 本单元的判别法是应对更难情形的工具。
If the terms $a_n\to 0$, the series must converge. · 若项$a_n\to 0$，则该级数必收敛。
Necessary but not sufficient — e.g. $\sum\tfrac1n$ diverges. · 必要但不充分条件 — 例如$\sum\tfrac1n$发散。
A series is a limit of partial sums, not just "the terms." Terms going to $0$ is necessary for convergence but not sufficient — $\sum\frac1n$ has terms $\to0$ yet diverges. Don't conclude convergence just because the terms shrink; you need one of the convergence tests.
级数是部分和的极限,而不只是"各项"。各项趋于 $0$ 对收敛是必要的但不充分——$\sum\frac1n$ 各项 $\to0$ 却发散。别因为各项缩小就断定收敛;你需要某个收敛判别法。
Does $1+\tfrac12+\tfrac14+\tfrac18+\cdots$ converge?
- Partial sums: $S_1=1$, $S_2=1.5$, $S_3=1.75$, $S_4=1.875,\dots$
- They climb toward $2$ and never pass it.
- The limit is $2$, so the series converges to $2$.
$1+\tfrac12+\tfrac14+\tfrac18+\cdots$ 收敛吗?
- 部分和:$S_1=1$、$S_2=1.5$、$S_3=1.75$、$S_4=1.875,\dots$
- 它们向 $2$ 攀升,永不越过。
- 极限是 $2$,所以级数收敛到 $2$。
An infinite series $\sum a_n$ is the limit of its partial sums $S_n=a_1+\cdots+a_n$: it converges to $L$ if $S_n\to L$ (finite), and diverges otherwise. Terms shrinking to $0$ is necessary but not sufficient for convergence.
无穷级数 $\sum a_n$ 是它的部分和 $S_n=a_1+\cdots+a_n$ 的极限:若 $S_n\to L$(有限)则收敛到 $L$,否则发散。各项缩向 $0$ 对收敛是必要但不充分的。