AP Calculus AB/BC · Unit 10 · Topic 10.8 · BC Only
Ratio Test for Convergence
Measure the geometric-scale behavior of a series by simplifying the absolute ratio of consecutive terms.
1. Topic Focus
Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.
This topic: Measure the geometric-scale behavior of a series by simplifying the absolute ratio of consecutive terms.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Use L<1 for absolute convergence, L>1 for divergence, and switch tests when L=1.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
BC-only topic. The Ratio Test compares the size of each term with the size of the preceding term. It is especially effective when factorials, exponentials, or products depending on \(n\) simplify after replacing \(n\) by \(n+1\).
Ratio Test
For a series \(\sum a_n\) with nonzero terms eventually, compute
- If \(0\le L<1\), then \(\sum a_n\) converges absolutely.
- If \(L>1\) or \(L=\infty\), then \(\sum a_n\) diverges.
- If \(L=1\), the Ratio Test is inconclusive.
Why a ratio below 1 gives convergence
If \(L<1\), choose a number \(r\) with \(L<r<1\). Eventually,
The tail of \(\sum|a_n|\) is bounded by a convergent geometric series. Therefore the original series converges absolutely.
Why a ratio above 1 gives divergence
If \(L>1\), the magnitudes eventually grow by a factor greater than \(1\). In particular, \(|a_n|\) cannot approach zero. The series therefore diverges by the nth term test.
The absolute value is essential
The test measures magnitudes, not alternating signs. For \(a_n=(-1)^n b_n\),
A result \(L<1\) proves the stronger conclusion of absolute convergence.
Algebra workflow
- Write a separate formula for \(a_{n+1}\).
- Form \(\left|a_{n+1}/a_n\right|\) before taking the limit.
- Turn division by \(a_n\) into multiplication by its reciprocal.
- Expand only the factorial factors needed for cancellation.
- Cancel common powers and factorials.
- Evaluate \(L\) and state the matching conclusion in words.
Factorial identities that save work
Do not expand an entire factorial. Expose only the new factors created by the index shift.
Recognize useful structures
- Polynomial over exponential: the polynomial ratio tends to \(1\), leaving the reciprocal exponential base.
- Exponential over factorial: the new factorial factor usually drives the ratio to \(0\).
- Factorial over exponential: the new factorial factor may make the ratio unbounded.
- Several factorials: shift every factorial carefully; \((2n)!\) gains two factors when \(n\) increases by \(1\).
- Power series: the ratio often produces a condition involving \(|x-a|\), which is developed further in Topic 10.13.
What \(L=1\) really means
The result \(L=1\) is not convergence and is not divergence. For every \(p>0\),
yet \(\sum1/n^p\) converges when \(p>1\) and diverges when \(0<p\le1\). A comparison, \(p\)-series test, Integral Test, or Alternating Series Test may be needed next.
When the ratio limit does not exist
The standard Ratio Test is also inconclusive when the limit of the absolute ratios does not exist. Do not average oscillating ratio values or select only a convenient subsequence; choose another convergence test.
Finite terms and zero terms
Changing finitely many terms does not change convergence. The usual ratio formula only needs to be defined eventually. If zeros continue to appear so that the quotient is repeatedly undefined, use a different test rather than forcing the calculation.
AP-style conclusion checklist
- Display the absolute ratio with \(a_{n+1}\) substituted correctly.
- Simplify enough to justify the limit.
- Compare \(L\) explicitly with \(1\).
- Say absolutely convergent when \(L<1\).
- Say divergent by the Ratio Test when \(L>1\).
- When \(L=1\), name another appropriate test instead of drawing a conclusion.
6. Detailed Worked Example and Error Check
Example 1: Polynomial divided by an exponential
Let \(a_n=n^2/5^n\). Then
Therefore \(\sum n^2/5^n\) converges absolutely.
Example 2: Exponential divided by a factorial
For \(a_n=3^n/n!\),
The series converges absolutely.
Example 3: Factorial growth dominates a fixed exponential
For \(a_n=n!/4^n\),
The series diverges. Its terms eventually grow and therefore fail the zero-term requirement.
Example 4: Alternating factorial expression
For \(a_n=(-1)^n(n!)^2/(2n)!\), absolute values remove the sign:
Because \(1/4<1\), the series converges absolutely.
Example 5: A ratio involving the number e
For \(a_n=n^n/n!\),
The series diverges by the Ratio Test.
Example 6: The harmonic series gives L = 1
The Ratio Test is inconclusive. The harmonic series diverges, but that conclusion comes from a different result.
Example 7: A convergent series also gives L = 1
Again the Ratio Test is inconclusive. The series converges because it is a \(p\)-series with \(p=2\).
Example 8: Two factorials in the denominator
For \(a_n=n!/(2n)!\),
The series converges absolutely.
Example 9: Factorial divided by a variable power
For \(a_n=n!/n^n\),
The series converges absolutely.
Example 10: A power-series preview
For a fixed real \(x\), let \(a_n=x^n/n!\). Then
Thus \(\sum x^n/n!\) converges absolutely for every real \(x\).
Common errors
- Forgetting the absolute value around \(a_{n+1}/a_n\).
- Using \(a_n/a_{n+1}\) but applying the standard conclusions unchanged.
- Replacing only some occurrences of \(n\) when constructing \(a_{n+1}\).
- Writing \((2n+2)!=(2n+2)(2n)!\) and omitting \(2n+1\).
- Expanding factorials completely and creating avoidable algebra errors.
- Concluding convergence rather than absolute convergence when \(L<1\).
- Concluding divergence when \(L=1\).
- Assuming the terms approach zero merely because a ratio was formed.
- Ignoring that an ordinary ratio limit does not exist.
- Using the Ratio Test on a simple rational-power series when a \(p\)-series or comparison argument is clearer.
7. AP Reasoning Routine
Check the nth-term condition first, match the series structure to a justified test, state convergence type, and test power-series endpoints separately.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use the Ratio Test when it gives a conclusion. If it is inconclusive, identify a suitable next test.
(a) \(\sum_{n=1}^{\infty}n^3/4^n\).
(b) \(\sum_{n=1}^{\infty}5^n/n!\).
(c) \(\sum_{n=1}^{\infty}n!/10^n\).
(d) \(\sum_{n=1}^{\infty}(-1)^n n^2/3^n\).
(e) \(\sum_{n=1}^{\infty}(n!)^2/(3n)!\).
(f) \(\sum_{n=1}^{\infty}2^n/n^4\).
(g) \(\sum_{n=1}^{\infty}1/n^3\).
(h) \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\).
(i) \(\sum_{n=1}^{\infty}n!/(2n)!\).
(j) Explain why \(L<1\) proves more than ordinary convergence.
(k) Suppose \(|a_{n+1}/a_n|\) alternates between values approaching \(1/2\) and \(3/2\). What does the standard Ratio Test prove?
(l) Write a complete AP-style justification for \(\sum_{n=1}^{\infty}(-1)^n7^n/n!\).
Check the solution
(a) \(L=\lim((n+1)/n)^3(1/4)=1/4<1\), so the series converges absolutely.
(b) \(L=\lim5/(n+1)=0\), so the series converges absolutely.
(c) \(L=\lim(n+1)/10=\infty\), so the series diverges.
(d) \(L=\lim((n+1)/n)^2(1/3)=1/3<1\), so it converges absolutely.
(e) \(L=\lim (n+1)^2/((3n+3)(3n+2)(3n+1))=0\), so it converges absolutely.
(f) \(L=\lim2(n/(n+1))^4=2>1\), so the series diverges.
(g) \(L=1\), so the Ratio Test is inconclusive. The \(p\)-series test gives convergence because \(p=3>1\).
(h) \(L=1\), so the Ratio Test is inconclusive. The Alternating Series Test gives convergence, while the absolute harmonic series diverges.
(i) \(L=\lim(n+1)/((2n+2)(2n+1))=0\), so the series converges absolutely.
(j) The ratio comparison is applied to \(\sum|a_n|\). Its convergence proves absolute convergence, which implies convergence of \(\sum a_n\).
(k) The absolute-ratio limit does not exist, so the standard Ratio Test is inconclusive. Another test is required.
(l) Let \(a_n=(-1)^n7^n/n!\). Then \(\lim|a_{n+1}/a_n|=\lim7/(n+1)=0<1\). Therefore the series converges absolutely by the Ratio Test.