Ratio Test for Series Convergence: How to Use It (Examples)

The ratio test decides whether an infinite series \( \sum a_n \) converges by looking at how fast its terms shrink. Compute

$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$

• If \( L < 1 \): the series converges (absolutely).
• If \( L > 1 \) (or \( L = \infty \)): the series diverges.
• If \( L = 1 \): the test is inconclusive; use another test.

This guide shows why the test works, a step-by-step routine, worked examples from easy to exam-level (including radius and interval of convergence), and the errors that most often cost points.

Why it works

If \( L < 1 \), then eventually each term is less than about \( L \) times the previous one, so the series is dominated by a geometric series with ratio below 1, which converges.

Think of the ratio as a “shrink factor” between neighboring terms. A geometric series has the same shrink factor \( r \) at every step, and it converges exactly when \( |r| < 1 \). The ratio test asks whether your series eventually behaves like a geometric series with a shrink factor below 1 or above 1.

A proof sketch

Case \( L < 1 \). Pick a number \( r \) with \( L < r < 1 \), for example halfway between them. Because the ratios approach \( L \), there is an index \( N \) after which every ratio is below \( r \). Then

$$|a_{N+1}| \le r|a_N|, \quad |a_{N+2}| \le r^2|a_N|, \quad \ldots$$

and in general \( |a_{N+k}| \le |a_N|\,r^k \). The right side is a convergent geometric series, so by comparison \( \sum|a_n| \) converges. The finitely many terms before \( N \) don’t affect convergence.

Case \( L > 1 \). Eventually every ratio exceeds 1, so each term is larger in absolute value than the one before. The terms cannot approach 0, and a series whose terms don’t go to 0 diverges.

Case \( L = 1 \). The comparison with a geometric series breaks down, because a shrink factor that creeps toward 1 can go either way. Example 4 shows both outcomes.

How to apply the ratio test

1. Write \( a_n \) and then \( a_{n+1} \) by replacing every \( n \) with \( n + 1 \).
2. Form \( \frac{a_{n+1}}{a_n} \) and flip the bottom fraction to multiply.
3. Cancel: \( \frac{(n+1)!}{n!} = n + 1 \) and \( \frac{c^{n+1}}{c^n} = c \).
4. Take the absolute value and the limit as n goes to infinity.
5. Compare \( L \) with 1 and state the conclusion.

When to use it

The ratio test shines when terms contain factorials \( n! \) or exponentials \( c^n \), because those simplify beautifully in a ratio.

Example 1: n / 2ⁿ

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} \to \frac12$$

\( L = \frac12 < 1 \), so \( \sum\frac{n}{2^n} \) converges (its sum is 2). The factor \( n \) grows, but only polynomially; the halving from \( 2^n \) wins.

Example 2: 3ⁿ / n!

$$\frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac{3}{n+1} \to 0$$

\( L = 0 < 1 \): converges. Factorials beat exponentials, which is why the series for \( e^x = \sum\frac{x^n}{n!} \) converges for every \( x \).

Example 3: n! / 10ⁿ

The ratio is \( \frac{n+1}{10} \to \infty \): diverges. The first few terms shrink, but once \( n \) passes 9 each term is bigger than the last. Only the long-run behavior matters.

Example 4: the inconclusive case

For \( \sum\frac{1}{n^2} \): \( \frac{n^2}{(n+1)^2} \to 1 \). For \( \sum\frac1n \): also \( \to 1 \). But the first converges (to \( \frac{\pi^2}{6} \)) and the second diverges. \( L = 1 \) tells you nothing. For these, use the p-series test: \( \sum\frac{1}{n^p} \) converges exactly when \( p > 1 \) (compare improper integrals).

Example 5: alternating signs

Test \( \sum(-1)^n\frac{n^3}{5^n} \). The absolute value removes the sign:

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^3}{n^3}\cdot\frac15 \to \frac15$$

\( L = \frac15 < 1 \), so the series converges absolutely. You don’t need the alternating series test here; the ratio test handled the signs by ignoring them.

Radius of convergence of a power series

For \( \sum c_nx^n \), apply the ratio test with \( x \) included, then solve \( L < 1 \) for \( x \).

Example: \( \sum\frac{x^n}{n\,2^n} \):

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1}\cdot\frac{|x|}{2} \to \frac{|x|}{2}$$

Converges when \( |x| < 2 \), so the radius of convergence is \( R = 2 \). (The endpoints need separate checks: it diverges at \( x = 2 \) and converges at \( x = -2 \).)

Two extremes are worth knowing. For \( \sum\frac{x^n}{n!} \) the ratio is \( \frac{|x|}{n+1} \to 0 \) for every \( x \), so \( R = \infty \). For \( \sum n!\,x^n \) the ratio is \( (n+1)|x| \to \infty \) unless \( x = 0 \), so \( R = 0 \).

Example 6 (exam-level): interval of convergence

Find the interval of convergence of \( \sum_{n=1}^{\infty}\frac{(2x - 1)^n}{n^2} \).

Ratio test. The ratio is \( |2x - 1|\cdot\frac{n^2}{(n+1)^2} \to |2x - 1| \). Convergence needs \( |2x - 1| < 1 \), which means \( 0 < x < 1 \). The center is \( \frac12 \) and the radius is \( \frac12 \).

Endpoints. The ratio test says nothing at \( L = 1 \), so test them directly:

• At \( x = 1 \): the series is \( \sum\frac{1}{n^2} \), a convergent p-series.
• At \( x = 0 \): the series is \( \sum\frac{(-1)^n}{n^2} \), which converges absolutely.

The interval of convergence is \( [0, 1] \). Forgetting the endpoint step is the single most common way to lose points on this question type.

Quick test selection guide

Terms look like	Try
\( n! \), \( c^n \)	ratio test
\( (\ldots)^n \)	root test
\( \frac{1}{n^p} \)-like	p-series / comparison
alternating signs	alternating series test
terms don’t go to 0	\( n \)th-term test: diverges

Common mistakes

• Flipping the ratio. It is \( \frac{a_{n+1}}{a_n} \), the new term over the old one. Upside down, a convergent series looks divergent.
• Mishandling factorials. \( (n+1)! = (n+1)\cdot n! \), so \( \frac{n!}{(n+1)!} = \frac{1}{n+1} \), not \( \frac1n \) or 1.
• Concluding from \( L = 1 \). “The limit is 1, so it converges” is wrong. Switch to a comparison, integral or p-series test.
• Thinking \( L \) is the sum. The ratio test only tells you whether the series converges. In Example 1, \( L = \frac12 \) but the sum is 2.
• Skipping endpoints. For power series, \( L = 1 \) exactly at the endpoints, so they always need a separate test.

Where it’s used

The ratio test is the workhorse for power series. Whenever you meet a Taylor series, the first question is where it converges, and the ratio test answers it in a few lines. That’s how you know the series for \( e^x \), \( \sin x \) and \( \cos x \) work for every \( x \), while the series for \( \ln(1 + x) \) stops at a radius of 1.

It also appears whenever a quantity grows or shrinks by a changing factor from one step to the next: probabilities in repeated trials, terms of recursive sequences, and error estimates for numerical methods. Some ratio limits need extra tools. For \( \sum\frac{n!}{n^n} \), the ratio simplifies to \( \left(\frac{n}{n+1}\right)^n \), and finding its limit of \( \frac1e \) takes logarithms and L’Hôpital’s rule.

Series convergence calculator

Enter \( a_n \) using n. The gadget computes partial sums, estimates the ratio-test limit and applies the \( n \)th-term test. The series convergence calculator page offers the same tool full width, and the limit calculator can evaluate a tricky ratio limit.

Practice problems

Try these before checking the answers.

1. \( \sum\frac{n^2}{3^n} \)
2. \( \sum\frac{2^n}{n!} \)
3. \( \sum\frac{n!}{n^n} \)
4. \( \sum\frac{n^5}{2^n} \)
5. \( \sum\frac{5^n}{n^2} \)
6. The radius of convergence of \( \sum\frac{x^n}{n!} \)

Answers: (1) \( L = \tfrac13 \), converges; (2) \( L = 0 \), converges; (3) \( L = \tfrac1e \), converges; (4) \( L = \tfrac12 \), converges; (5) \( L = 5 \), diverges; (6) \( R = \infty \).

FAQ

Does the ratio test prove absolute convergence?

Yes. When \( L < 1 \), \( \sum|a_n| \) converges, so \( \sum a_n \) does too.

Can I use the ratio test on p-series?

You can, but it always gives \( L = 1 \), so it never decides them.

What’s the difference between the ratio test and the root test?

The root test uses \( \lim\sqrt[n]{|a_n|} \) with the same three outcomes. It’s easier when the whole term is raised to the \( n \)th power; the ratio test is easier with factorials.

Why does the ratio test use absolute values?

Signs don’t affect how fast terms shrink in size. Taking absolute values lets one test cover alternating series too, and it is why a pass proves absolute convergence.

How does the ratio test relate to Taylor series?

It is the standard way to find the radius of convergence of a Taylor series, which tells you for which \( x \) the series equals its function.

Further reading

• Ratio Test (Paul’s Online Math Notes) — a factorial review, six worked examples and a full proof.
• Ratio and Root Tests (OpenStax Calculus Volume 2) — both tests side by side with checkpoint exercises.