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\(\left| \frac{a_{n+1}}{a_n} \right|\)

Ratio Test for
Convergence

Master the Ratio Test to determine whether an infinite series converges or diverges by examining the limit of consecutive term ratios.

AP Calculus BC

Why Does Series Convergence Matter?

When we add infinitely many terms, the sum might settle to a finite value (converge) or grow without bound (diverge). Knowing which happens is essential for Taylor series, power series, and many applications in physics and engineering.

The Consecutive Term Ratio

Given a series \(\displaystyle\sum_{n=1}^{\infty} a_n\), the ratio of consecutive terms is:

If consecutive terms shrink fast enough, the series converges. If they grow, it diverges. The Ratio Test formalizes this idea by taking the limit of this ratio.

Visual Intuition

\(n\)
\(|a_n|\)
 Converges: ratio < 1 Diverges: ratio > 1 1 2 3 4 5 6 1 2 3 4

The Ratio Test: Formal Statement

Given \(\displaystyle\sum_{n=1}^{\infty} a_n\) with \(a_n \neq 0\), let:

\(L < 1\)
Series converges absolutely
\(L > 1\)
Series diverges
\(L = 1\)
Test is inconclusive

Why \(L = 1\) Is Inconclusive

Two classic series both give \(L = 1\), yet one converges and the other diverges:

\(\displaystyle\sum \frac{1}{n^2}\)
Converges (p-series, \(p=2\))
\(L = \lim \frac{n^2}{(n+1)^2} = 1\)
\(\displaystyle\sum \frac{1}{n}\)
Diverges (harmonic series)
\(L = \lim \frac{n}{n+1} = 1\)

Ratio Test vs Root Test

Ratio Test Root Test
Limit \(\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|\) \(\displaystyle\lim\sqrt[n]{|a_n|}\)
Best for Factorials (\(n!\)), products Powers of \(n\), \(n^n\)
Criteria \(L<1\): conv, \(L>1\): div Same criteria
Strength Easier to compute with \(n!\) Handles \(a_n = f(n)^n\) better

When to Use the Ratio Test

Factorials
Terms like \(\frac{n!}{3^n}\) or \(\frac{2^n}{n!}\) simplify beautifully with the ratio.
Exponentials
Terms like \(\frac{5^n}{n^3}\) where consecutive ratios cancel cleanly.
Products
Complex products where \(\frac{a_{n+1}}{a_n}\) collapses to a simple expression.

How to Apply the Ratio Test

1
Write \(a_n\) explicitly. Identify the general term of the series.
2
Form the ratio \(\left|\dfrac{a_{n+1}}{a_n}\right|\). Replace every \(n\) with \(n+1\) in the numerator.
3
Simplify. Cancel common factors: factorials, powers, and products.
4
Take the limit as \(n \to \infty\). This gives you \(L\).
5
Conclude: \(L < 1\) → converges, \(L > 1\) → diverges, \(L = 1\) → inconclusive.

Example 1: \(\displaystyle\sum \frac{n!}{3^n}\)

1
General term: \(a_n = \dfrac{n!}{3^n}\)
2
Ratio: \(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n+1)!}{3^{n+1}} \cdot \dfrac{3^n}{n!}\)
3
Simplify: \(= \dfrac{(n+1) \cdot n!}{3 \cdot 3^n} \cdot \dfrac{3^n}{n!} = \dfrac{n+1}{3}\)
4
Limit: \(L = \lim_{n\to\infty} \dfrac{n+1}{3} = \infty\)

Example 2: \(\displaystyle\sum \frac{2^n}{n!}\)

1
General term: \(a_n = \dfrac{2^n}{n!}\)
2
Ratio: \(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{2^{n+1}}{(n+1)!} \cdot \dfrac{n!}{2^n}\)
3
Simplify: \(= \dfrac{2 \cdot 2^n}{(n+1) \cdot n!} \cdot \dfrac{n!}{2^n} = \dfrac{2}{n+1}\)
4
Limit: \(L = \lim_{n\to\infty} \dfrac{2}{n+1} = 0\)

Example 3: \(\displaystyle\sum \frac{n^3 \cdot 5^n}{n!}\)

1
Ratio: \(\dfrac{(n+1)^3 \cdot 5^{n+1}}{(n+1)!} \cdot \dfrac{n!}{n^3 \cdot 5^n}\)
2
Simplify: \(= \dfrac{(n+1)^3}{n^3} \cdot \dfrac{5}{n+1} = \dfrac{5(n+1)^2}{n^3}\)
3
Limit: \(L = \lim_{n\to\infty} \dfrac{5(n+1)^2}{n^3} = \lim_{n\to\infty} \dfrac{5}{n} \cdot \left(\dfrac{n+1}{n}\right)^2 = 0\)

Visualizing the Ratio \(\left|\frac{a_{n+1}}{a_n}\right|\)

L = 1
\(n\)
\(\left|\frac{a_{n+1}}{a_n}\right|\)
 1 0 → 0 → ∞
\(\frac{2^n}{n!}\) (converges)
\(\frac{n!}{3^n}\) (diverges)
 1 2 3 4 5 6 7 8

Absolute vs Conditional Convergence

The Ratio Test uses absolute values, so when \(L < 1\) it proves absolute convergence. This is stronger than conditional convergence.

Absolute Convergence
\(\sum |a_n|\) converges, which implies \(\sum a_n\) also converges.
Conditional Convergence
\(\sum a_n\) converges but \(\sum |a_n|\) diverges. The Ratio Test cannot detect this.

Common Mistakes to Avoid

✕ Forgetting Absolute Value
The Ratio Test requires \(\left|\frac{a_{n+1}}{a_n}\right|\). Omitting absolute values leads to incorrect conclusions for alternating series.
✕ Concluding at \(L = 1\)
When \(L = 1\), the test is inconclusive: not convergent, not divergent. You must use a different test.
✕ Factorial Algebra Errors
Remember: \((n+1)! = (n+1) \cdot n!\). Do not write \((n+1)! = n! + 1\) or expand the entire factorial.

Quick Reference Summary

Series Type Ratio \(L\) Result
\(\displaystyle\sum \frac{2^n}{n!}\) \(0\) Converges
\(\displaystyle\sum \frac{n!}{3^n}\) \(\infty\) Diverges
\(\displaystyle\sum \frac{n^3 \cdot 5^n}{n!}\) \(0\) Converges
\(\displaystyle\sum \frac{n^2}{2^n}\) \(\frac{1}{2}\) Converges
\(\displaystyle\sum \frac{1}{n^2}\) \(1\) Inconclusive
\(\displaystyle\sum \frac{3^n}{n^5}\) \(3\) Diverges

Question 1 of 5

Apply the Ratio Test to \(\displaystyle\sum_{n=1}^{\infty} \frac{n!}{5^n}\). What is the value of \(L\)?

Question 2 of 5

For \(\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n!}\), the Ratio Test gives:

Question 3 of 5

Which of the following series should you NOT test with the Ratio Test?

Question 4 of 5

Apply the Ratio Test to \(\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}\). What is \(L\)?

Question 5 of 5

When the Ratio Test gives \(L = 1\), which statement is true?