Calculus II · Unit 4A · lesson

The Ratio Test

Detect geometric decay in factorial and exponential series, prove the Ratio Test mechanism, and recognize the inconclusive boundary case.

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Section overview

Alternating, absolute, ratio, and root tests

What this section is building

Detect geometric decay in factorial and exponential series, prove the Ratio Test mechanism, and recognize the inconclusive boundary case.

Notice

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

Decide

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

Avoid

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

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 $\lim|a_{n+1}/a_n|$ .

Your work stays on this device. No account or AI grader is used.

Show hint

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

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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 $L<1$ .

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

The Ratio Test

Explanation

Consecutive ratios expose multiplicative growth

Bridge

Look at the factor relating neighboring terms

Read this graph as text

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. Successive term bars shrinking by an approximately fixed factor. Preserve the misconception control described in the lesson and storyboard.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in a ratio below one creates geometric domination; color is never the only cue.

Why it matters: Successive term bars shrinking by an approximately fixed factor.

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 $\lim|a_{n+1}/a_n|$ .

Your work stays on this device. No account or AI grader is used.

Show hint

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

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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 $L<1$ .

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

The Ratio Test

Related resources

Original instruction with traceable references.

The Ratio Test

Related resources

Original instruction with traceable references.