Calculus II · Unit 4A · lesson

The Ratio Test

Concept

Learning objectives

use ratios of consecutive terms to detect factorial and exponential behavior.

The Ratio Test

Explanation

Consecutive ratios expose multiplicative growth

Factorials and exponentials are awkward for comparison term by term but simple under division by the preceding term. The Ratio Test measures the eventual multiplication factor in term magnitudes. If that factor is below one, the series behaves like a convergent geometric series; if above one, terms fail to shrink fast enough.

A ratio limit equal to one is inconclusive, not evidence of convergence. This happens for many $p$ -series, so another test must take over. Always apply the ratio to absolute values and simplify before taking the limit. For power series, the ratio test later becomes the standard route to a radius of convergence.

Bridge

Look at the factor relating neighboring terms

The Ratio Test is designed for factorials, exponentials, and products whose neighboring terms simplify dramatically. If $|a_{n+1}/a_n|\to L<1$ , then the tail eventually shrinks by a fixed factor smaller than one and is dominated by a geometric series.

The boundary $L=1$ is genuinely inconclusive, not a weak hint. Both the harmonic series and $\sum1/n^2$ have ratio limit one, yet one diverges and the other converges. When the test returns one, choose a different method.

**A ratio below one creates geometric domination**
Successive term magnitudes are shown decreasing by a factor bounded above by a fixed number r less than one.
A ratio below one creates geometric domination. Successive term bars shrinking by an approximately fixed factor.

Proof idea

A limit below one yields an actual geometric bound

Choose $r$ with $L<r<1$ . Eventually $|a_{n+1}|\le r|a_n|$ , so repeated application gives $|a_{N+k}|\le |a_N|r^k$ . The tail is bounded by a convergent geometric series.

Concept

Ratio Test

Let

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 $L=\infty$ , it diverges. If $L=1$ , the test is inconclusive.

Guided walkthrough

Factorial in the denominator

For $a_n=3^n/n!$ ,

\left|\frac{a_{n+1}}{a_n}\right| =\frac{3^{n+1}}{(n+1)!}\frac{n!}{3^n} =\frac3{n+1}\to0.

Therefore $\sum3^n/n!$ converges absolutely.

Worked example

Factorials collapse under neighboring ratios

Test

\sum_{n=1}^{\infty}\frac{3^n}{n!}.

Let $a_n=3^n/n!$ . Then

\left|\frac{a_{n+1}}{a_n}\right| =\frac{3^{n+1}}{(n+1)!}\frac{n!}{3^n} =\frac3{n+1}\to0.

Since $0<1$ , the series converges absolutely.

Common mistake

L equals one means stop using this test

A ratio limit of one gives no conclusion. Do not label it “barely convergent” or “probably divergent.” Switch tests.

Interactive checku4a-ratio_test-01

For $a_n=3^n/n!$ , find $\lim|a_{n+1}/a_n|$ .

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Show hint

Cancel $3^n$ and $n!$ .

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Exercise

Apply the Ratio Test to $\sum n!/4^n$ .

Exercise

Apply it to $\sum n^2/5^n$ .

Exercise

Show why the Ratio Test is inconclusive for $\sum1/n^2$ .

Exercise

Explain the geometric-series intuition behind $L<1$ .

After the explanation

Use the section idea

Reading lens

Separate sign behavior from magnitude, then use ratios or roots when powers and factorials dominate.

Mental model

Absolute convergence controls magnitude strongly enough to imply convergence; conditional convergence relies on cancellation.

Decision

Test absolute values first when practical, use the alternating-series hypotheses explicitly, and reserve ratio or root tests for matching algebraic structure.

Common trap

Calling any alternating-looking series convergent or treating a ratio/root limit of one as a verdict.

Check yourself

Can you state whether convergence is absolute, conditional, divergent, or still undecided?

Source & rights

Original instruction with traceable references.

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