Harmonic Series and p-Series
Introduced| English |
|---|
| harmonic series/hɑːˈmɒnɪk ˈsɪəriːz/ |
| p-series/piː ˈsɪəriːz/ |
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.
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 $p=1$.
How fast do the terms shrink?
A p-series converges only if its terms $\tfrac1{n^p}$ shrink fast enough — the cutoff is exactly $p>1$.
The p-series $\sum\tfrac1{n^p}$ converges exactly when...
Converges iff $p>1$.
The convergence cutoff for a p-series is exactly $p=$ ____.
Converge for $p>1$, diverge for $p\le1$.
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.
The harmonic series $\sum\tfrac1n$ ($p=1$)...
$p=1$ diverges, even though terms $\to0$.
The harmonic series shows terms going to $0$ does not guarantee convergence.
Its terms $\to0$ yet it diverges.
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.
Select all convergent series.
$p=2,3>1$ converge; $p=\tfrac12,1$ diverge.
$\sum\tfrac1{\sqrt n}$ is a p-series with $p=\tfrac12$. It...
$p=\tfrac12\le1$ → diverges.
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 $1$.
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.
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.