The Ratio Test
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 5.6: “Ratio and Root Tests” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/5-6-ratio-and-root-tests |
| Textbook used in class | Stewart, Calculus, Section 11.6: “Absolute Convergence and the Ratio and Root Tests” (Examples 1, 2, 3) |
Opening Scenario
Factorials grow faster than any exponential: $n!$ eventually dwarfs $10^n$. The Ratio Test detects this by comparing each term to the next. If the ratio $|a_{n+1}/a_n|$ shrinks toward a limit less than $1$, each term is a shrinking fraction of the previous one -- the series behaves like a geometric series with ratio less than $1$ and must converge.
Quick Reference
Ratio Test. Let $\displaystyle L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$.
- If $L < 1$: $\sum a_n$ converges absolutely.
- If $L > 1$ (or $L = \infty$): $\sum a_n$ diverges.
- If $L = 1$: inconclusive -- the test gives no information.
Best for: terms involving factorials ($n!$), exponentials ($r^n$), or products of these. Fails for: $p$-series and any series where the ratio limit is $1$ (polynomial expressions in $n$).
Key Concepts
1. Why It Works
If $L < 1$, choose $r$ with $L < r < 1$. For large $n$, $|a_{n+1}| < r|a_n|$, so the terms shrink geometrically. The tail of the series is bounded above by a convergent geometric series with ratio $r$.
2. Computing the Ratio
Example 1. Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^n}{n!}$ converge? (Stewart 11.6, Example 1.)
Wait -- that one diverges, but let me use the standard textbook example.
Example 1. Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n}$ converge? (Stewart 11.6, Example 1.)
Compute: \[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} = \frac{(n+1)^3}{3 n^3} = \frac{1}{3}\left(\frac{n+1}{n}\right)^3 \to \frac{1}{3} \cdot 1 = \frac{1}{3}. \]
$L = 1/3 < 1$: the series converges absolutely.
Boxed answer: Converges absolutely.
Example 2. Does $\displaystyle\sum_{n=0}^{\infty} \frac{n!}{100^n}$ converge? (Stewart 11.6, Example 2 type.)
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{100^{n+1}} \cdot \frac{100^n}{n!} = \frac{n+1}{100} \to \infty. \]
$L = \infty > 1$: the series diverges.
Boxed answer: Diverges.
Example 3. Apply the ratio test to $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$.
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1} \to 1. \]
$L = 1$: inconclusive. (The harmonic series actually diverges, but the ratio test cannot tell you that.)
3. The $L = 1$ Case
When $L = 1$, the ratio test says absolutely nothing. You must use a different test. Any $p$-series or series resembling a polynomial ratio will give $L = 1$ under the ratio test.
concluding convergence from $L = 1$. Both the convergent series $\sum 1/n^2$ and the divergent series $\sum 1/n$ give ratio-test limit $L = 1$. When the ratio test is inconclusive, do not guess based on the limit value -- switch to a comparison test, integral test, or $p$-series test.
Common Errors Summary
| Error | Correction |
|---|---|
| Forgetting absolute value in $|a_{n+1}/a_n|$ | The ratio uses absolute values, so sign cancels |
| Concluding convergence when $L = 1$ | $L = 1$ is inconclusive; use another test |
| Applying ratio test to a $p$-series | Always gives $L = 1$; use $p$-series test directly |
| Simplification error with factorials | $\frac{(n+1)!}{n!} = n+1$; cancel all common factors |
Common Misconceptions
a ratio test limit of $L = 1$ means the series converges.
This is the limit-equals-function-value error. When $L = 1$, the ratio test is inconclusive: it provides no information about convergence or divergence. Both the convergent series $\sum 1/n^2$ and the divergent harmonic series $\sum 1/n$ produce $L = 1$ under the ratio test. Concluding convergence from $L = 1$ is always wrong; a different test must be used.
the ratio test is applicable to all series, including $p$-series.
This is the concept-image-conflicts-definition error about the test’s effective range. For any $p$-series $\sum 1/n^p$, the ratio test gives $L = \lim n^p/(n+1)^p = 1$, which is always inconclusive. The ratio test is designed for terms with factorials or pure exponentials, where the ratio $|a_{n+1}/a_n|$ simplifies to something other than $1$. Applying it to polynomial or rational expressions wastes effort.
Leveled Practice
Level 1
Problem 1. Use the ratio test on $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!}$.
Show answer
$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{2^{n+1}}{(n+1)!} \cdot \dfrac{n!}{2^n} = \dfrac{2}{n+1} \to 0$.
$L = 0 < 1$: converges absolutely.
Problem 2. Use the ratio test on $\displaystyle\sum_{n=1}^{\infty} \frac{n^2 \cdot 3^n}{n!}$.
Show answer
$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n+1)^2 \cdot 3^{n+1}}{(n+1)!} \cdot \dfrac{n!}{n^2 \cdot 3^n} = \dfrac{3(n+1)^2}{(n+1) n^2} = \dfrac{3(n+1)}{n^2} \to 0$.
Converges absolutely.
Mastery Checklist
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