Harmonic Series and p-Series · 调和级数与p-级数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| harmonic series/hɑːˈmɒnɪk ˈsɪəriːz/ | 调和级数 | tiáo hé jí shù |
| p-series/piː ˈsɪəriːz/ | p-级数 | p- jí shù |
A whole family of test cases
- Two related series come up so often they get names — and a simple convergence rule.
- The harmonic series 调和级数 is $\sum\tfrac1n=1+\tfrac12+\tfrac13+\cdots$.
- The p-series p-级数 generalizes it: $\sum\tfrac1{n^p}$ for a fixed power $p$.
- One clean threshold on $p$ decides convergence for the whole family.
一整族测试案例
- 两个相关的级数出现得如此频繁,以至于有了名字——以及一条简单的收敛规则。
- 调和级数是 $\sum\tfrac1n=1+\tfrac12+\tfrac13+\cdots$。
- p-级数把它推广:对固定幂 $p$,$\sum\tfrac1{n^p}$。
- 一个关于 $p$ 的干净阈值就为整族判定收敛。
The p-series rule
- $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}$ converges if $p>1$ and diverges if $p\le 1$.
- Bigger $p$ makes the terms shrink faster — fast enough to converge once $p$ passes $1$.
- This follows directly from the Integral Test on $\int_1^\infty x^{-p}\,dx$.
- Memorize the threshold: the cutoff is exactly $p=1$.
p-级数规则
- $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}$ 在 $p>1$ 时收敛,在 $p\le 1$ 时发散。
- 更大的 $p$ 让各项缩得更快——一旦 $p$ 超过 $1$,快到足以收敛。
- 这直接由对 $\int_1^\infty x^{-p}\,dx$ 的积分判别法得出。
- 记住阈值:分界线恰好是 $p=1$。
How fast do the terms shrink? · 项衰减的速度有多快？
A p-series converges only if its terms $\tfrac1{n^p}$ shrink fast enough — the cutoff is exactly $p>1$. · p-级数仅当其项$\tfrac1{n^p}$衰减足够快时才收敛 — 阈值恰好是$p>1$。
The p-series $\sum\tfrac1{n^p}$ converges exactly when... · p-级数$\sum\tfrac1{n^p}$当且仅当...时收敛。
Converges iff $p>1$. · 当且仅当$p>1$时收敛。
The convergence cutoff for a p-series is exactly $p=$ ____. · p-级数的收敛阈值恰好是$p=$ ____。
Converge for $p>1$, diverge for $p\le1$. · 当$p>1$时收敛，当$p\le1$时发散。
The harmonic series is the borderline
- The harmonic series is the $p=1$ case — and $p=1$ diverges.
- Its terms $\tfrac1n\to0$, yet the sum grows without bound (slowly, but forever).
- It's the classic "terms shrink but the series still diverges" example.
- $p=1$ sits exactly on the wrong side of the cutoff.
调和级数是临界点
- 调和级数是 $p=1$ 的情形——而 $p=1$ 发散。
- 它的各项 $\tfrac1n\to0$,可是和无界增长(慢,却永不停)。
- 它是经典的"各项缩小但级数仍发散"的例子。
- $p=1$ 恰好落在分界线错误的一侧。
The harmonic series $\sum\tfrac1n$ ($p=1$)... · 调和级数$\sum\tfrac1n$ ($p=1$)...
$p=1$ diverges, even though terms $\to0$. · $p=1$发散，尽管项$\to0$。
The harmonic series shows terms going to $0$ does not guarantee convergence. · 调和级数表明项趋于$0$并不能保证收敛。
Its terms $\to0$ yet it diverges. · 其项$\to0$但它发散。
Instant convergence checks
- Recognize a p-series and you skip all the work — just compare $p$ with $1$.
- $\sum\tfrac1{n^2}$: $p=2>1$ → converges. $\sum\tfrac1{\sqrt n}=\sum\tfrac1{n^{1/2}}$: $p=\tfrac12\le1$ → diverges.
- These are also the standard comparison series for the comparison tests (next lesson).
- Knowing p-series cold speeds up a huge fraction of series problems.
即时收敛检查
- 认出一个 p-级数,你就跳过所有工作——只需把 $p$ 与 $1$ 比较。
- $\sum\tfrac1{n^2}$:$p=2>1$ → 收敛。$\sum\tfrac1{\sqrt n}=\sum\tfrac1{n^{1/2}}$:$p=\tfrac12\le1$ → 发散。
- 这些也是比较判别法的标准比较级数(下一课)。
- 把 p-级数烂熟于心能加速一大部分级数问题。
Select all · 所有 convergent series. · 选择所有收敛级数。
$p=2,3>1$ converge; $p=\tfrac12,1$ diverge. · $p=2,3>1$收敛；$p=\tfrac12,1$发散。
$\sum\tfrac1{\sqrt n}$ is a p-series with $p=\tfrac12$. It... · $\sum\tfrac1{\sqrt n}$是一个p-级数，其$p=\tfrac12$。它...
$p=\tfrac12\le1$ → diverges. · $p=\tfrac12\le1$ → 发散。
The cutoff is strict: $\sum\tfrac1{n^p}$ converges only for $p>1$. The boundary case $p=1$ (harmonic) diverges — a very common trap, since its terms do go to $0$. And $\sum\tfrac1{\sqrt n}$ is $p=\tfrac12$, which diverges (not converges) — a fractional power below $1$.
分界是严格的:$\sum\tfrac1{n^p}$ 仅在 $p>1$ 时收敛。边界情形 $p=1$(调和)发散——一个很常见的陷阱,因为它的各项确实趋于 $0$。而 $\sum\tfrac1{\sqrt n}$ 是 $p=\tfrac12$,它发散(不是收敛)——一个小于 $1$ 的分数幂。
Classify each series.
- $\sum\tfrac1{n^3}$: $p=3>1$ → converges.
- $\sum\tfrac1{n}$: $p=1$ → diverges (harmonic).
- $\sum\tfrac1{n^{0.9}}$: $p=0.9\le1$ → diverges.
判断各级数。
- $\sum\tfrac1{n^3}$:$p=3>1$ → 收敛。
- $\sum\tfrac1{n}$:$p=1$ → 发散(调和)。
- $\sum\tfrac1{n^{0.9}}$:$p=0.9\le1$ → 发散。
A p-series $\sum\tfrac1{n^p}$ converges iff $p>1$; it diverges for $p\le1$. The harmonic series is the $p=1$ case, which diverges despite its terms going to $0$. Recognizing a p-series gives an instant convergence verdict.
p-级数 $\sum\tfrac1{n^p}$ 收敛当且仅当 $p>1$;它在 $p\le1$ 时发散。调和级数是 $p=1$ 的情形,尽管各项趋于 $0$ 仍发散。认出 p-级数就给出即时的收敛判定。