| English | 의미 |
|---|---|
| Ratio Test/ˈreɪʃɪəʊ test/ | 비율 판별법 |
Comparing a term to the one before it
- When terms involve factorials or $n$th powers, comparisons are awkward. Use the Ratio Test 比值判别法 instead.
- It looks at how each term compares to the previous one — the ratio $\big|\tfrac{a_{n+1}}{a_n}\big|$ — in the limit.
- A ratio settling below $1$ means terms shrink geometrically fast, forcing convergence.
- It's the go-to test for factorials and exponentials.
The test
- Compute $\displaystyle L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$.
- $L<1$: the series converges (absolutely).
- $L>1$ (or $L=\infty$): the series diverges.
- $L=1$: the test is inconclusive — use another test.
1보다 작은 비율은 수렴을 강제합니다.
연속 항의 비율이 $1$ 아래로 수렴하면, 항들은 기하급수적으로 빠르게 감소하여 — 급수는 수렴합니다.
비율 검정법에서, $L=\lim|a_{n+1}/a_n|<1$이면, 급수는...
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$L<1$ → 수렴함.
비율 검정법은 연속 항의 비율의 절댓값을 사용합니다.
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$L=\lim|a_{n+1}/a_n|$.
Factorials and powers love the ratio test
- Factorials telescope beautifully: $\dfrac{(n+1)!}{n!}=n+1$ — the ratio simplifies enormously.
- Same for $n$th powers of a constant: $\dfrac{r^{n+1}}{r^n}=r$.
- So series like $\sum\tfrac{2^n}{n!}$ or $\sum\tfrac{n!}{n^n}$ are natural ratio-test problems.
- Write $\tfrac{a_{n+1}}{a_n}$, cancel, and take the limit.
비율 검정법이 $L=1$을 주면, ...
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$L=1$은 아무런 정보를 주지 않습니다.
$a_n=\tfrac{2^n}{n!}$의 경우, $\dfrac{a_{n+1}}{a_n}=$
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$\frac{2^{n+1}/(n+1)!}{2^n/n!}=\frac{2}{n+1}$.
비율 검정법은 ...를 가진 급수에서 특히 잘 작동합니다.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
p-급수는 $L=1$(무용)을 제공하며, 계승/지수함수는 잘 정리된다.
The $L=1$ dead zone
- When $L=1$, the ratio test gives no information — the series could converge or diverge.
- This happens for p-series (e.g. $\sum\tfrac1{n^2}$ gives $L=1$), where you'd use the p-series rule instead.
- So the ratio test shines on factorials/exponentials but is useless on plain powers of $n$.
- Recognize $L=1$ and switch tools.
$\lim\tfrac{2}{n+1}=0<1$이므로, 급수 $\sum\tfrac{2^n}{n!}$...
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$L=0<1$ → 수렴함.
Take the absolute value of the ratio, and remember the three cases: $L<1$ converges, $L>1$ diverges, $L=1$ is inconclusive (don't guess — use another test). The ratio test is powerful for factorials/exponentials but gives $L=1$ (useless) on p-series like $\sum\tfrac1{n^p}$.
Does $\displaystyle\sum_{n=1}^{\infty}\dfrac{2^n}{n!}$ converge?
- $\dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}/(n+1)!}{2^n/n!}=\dfrac{2}{n+1}$.
- $\displaystyle L=\lim_{n\to\infty}\dfrac{2}{n+1}=0<1$.
- Since $L<1$, the series converges.
The Ratio Test: $L=\lim\big|\frac{a_{n+1}}{a_n}\big|$. $L<1$ → converges, $L>1$ → diverges, $L=1$ → inconclusive. Ideal for factorials and exponentials (the ratios simplify); useless (gives $L=1$) on p-series.