The nth Term Test for Divergence · n项判别法判定发散
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| nth Term Test/ˌenˌtiːˈeɪtʃ tɜːm test/ | 通项判别法 | tōng xiàng pàn bié fǎ |
A quick way to prove divergence
- Before running a fancy test, do the cheapest check: look at the terms.
- The nth Term Test 通项判别法 (or Divergence Test) uses the limit of the terms $a_n$.
- If the terms don't shrink to $0$, the series can't converge — it diverges immediately.
- It's the first thing to try, and it only ever proves divergence.
快速证明发散的方法
- 在动用复杂判别法之前,做最省事的检查:看各项。
- 项判别法(或发散判别法)使用各项 $a_n$ 的极限。
- 若各项不缩向 $0$,级数不可能收敛——立即发散。
- 它是首选,而且只能证明发散。
The test
- Compute $\displaystyle\lim_{n\to\infty} a_n$.
- If $\displaystyle\lim_{n\to\infty} a_n \neq 0$ (or the limit doesn't exist), the series $\sum a_n$ diverges.
- Intuition: if you keep adding pieces that don't shrink to $0$, the total can never settle.
- One limit computation, and you may be done.
判别法
- 计算 $\displaystyle\lim_{n\to\infty} a_n$。
- 若 $\displaystyle\lim_{n\to\infty} a_n \neq 0$(或极限不存在),级数 $\sum a_n$ 发散。
- 直觉:若你不断加上不缩向 $0$ 的块,总和永远无法安定。
- 一次极限计算,你可能就完成了。
Terms that don't reach zero · 不趋于零的项
If a series' terms level off at a nonzero value, the sum can never settle — the nth Term Test proves divergence. · 若级数的项稳定在非零值，其和永不可能收敛 — n项判别法证明发散。
If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$... · 若$\lim_{n\to\infty}a_n\neq 0$，则$\sum a_n$...
Nonzero term limit → divergence. · 非零项极限 → 发散。
It can never prove convergence
- If $\displaystyle\lim_{n\to\infty} a_n = 0$, the test is inconclusive — it tells you nothing.
- The series might converge or diverge; you must use another test.
- The harmonic series $\sum\tfrac1n$ has $a_n\to0$ but diverges — proof the test can't confirm convergence.
- So a "$0$" limit is a dead end for this test, not a green light.
它永远不能证明收敛
- 若 $\displaystyle\lim_{n\to\infty} a_n = 0$,判别法不确定——它什么都不告诉你。
- 级数可能收敛也可能发散;你必须用别的判别法。
- 调和级数 $\sum\tfrac1n$ 有 $a_n\to0$ 却发散——证明该判别法不能确认收敛。
- 所以"$0$"极限对这个判别法是死胡同,而非通行灯。
If $\lim_{n\to\infty}a_n=0$, the nth Term Test is... · 若$\lim_{n\to\infty}a_n=0$，n项判别法是...
It only proves divergence; a $0$ limit tells you nothing. · 它只能证明发散；$0$极限什么都不能告诉你。
The nth Term Test can prove a series converges. · n项判别法可以证明级数收敛。
It can only prove divergence. · 它只能证明发散。
The series $\sum\tfrac1n$ has $a_n\to0$ but diverges. This shows the nth Term Test... · 级数$\sum\tfrac1n$具有$a_n\to0$但发散。这表明n项判别法...
A $0$ limit doesn't guarantee convergence. · $0$极限不能保证收敛。
Where it fits in the strategy
- Always run the nth Term Test first — it's fast and can save you a hard test.
- Terms not $\to0$? Done: diverges.
- Terms $\to0$? Move on to a real convergence test (geometric, integral, comparison, ratio, alternating).
- It's a filter, not a full answer.
它在策略中的位置
- 永远先跑项判别法——它快,能省下一个难判别法。
- 各项不 $\to0$?完成:发散。
- 各项 $\to0$?转向真正的收敛判别法(几何、积分、比较、比值、交错)。
- 它是一个过滤器,不是完整答案。
For · 支持 $\sum\dfrac{n}{2n+1}$, $\lim a_n=\tfrac12$. So the series... · 对于$\sum\dfrac{n}{2n+1}$，$\lim a_n=\tfrac12$。所以该级数...
$\tfrac12\neq0$ → diverges. · $\tfrac12\neq0$ → 发散。
As a fast filter, run the nth Term Test ____ among the convergence tests. · 作为一个快速筛选器，应在收敛性检验中____运行n项判别法。
It's the cheapest divergence check. · 这是最廉价的发散检查。
The nth Term Test can only prove divergence. If $\lim a_n=0$, it is inconclusive — do not conclude convergence. The single most common misuse is "$a_n\to0$, so it converges" — wrong; $\sum\frac1n$ is the counterexample.
项判别法只能证明发散。若 $\lim a_n=0$,它不确定——不要断定收敛。最常见的误用是"$a_n\to0$,所以收敛"——错;$\sum\frac1n$ 就是反例。
Does $\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{2n+1}$ converge?
- Terms: $a_n=\dfrac{n}{2n+1}$. Limit: $\displaystyle\lim_{n\to\infty}\dfrac{n}{2n+1}=\dfrac12$.
- Since the limit is $\tfrac12\neq0$, the series diverges by the nth Term Test.
$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{2n+1}$ 收敛吗?
- 各项:$a_n=\dfrac{n}{2n+1}$。极限:$\displaystyle\lim_{n\to\infty}\dfrac{n}{2n+1}=\dfrac12$。
- 因为极限是 $\tfrac12\neq0$,由项判别法级数发散。
The nth Term Test: if $\lim_{n\to\infty}a_n\neq 0$ (or DNE), then $\sum a_n$ diverges. If $\lim a_n=0$, the test is inconclusive (it never proves convergence). Run it first as a fast divergence filter.
项判别法:若 $\lim_{n\to\infty}a_n\neq 0$(或不存在),则 $\sum a_n$ 发散。若 $\lim a_n=0$,判别法不确定(它永不证明收敛)。先跑它作为快速的发散过滤器。