AP Calculus AB/BC · Unit 10 · Topic 10.6 · BC Only
Comparison Tests for Convergence
Transfer convergence or divergence from a known positive benchmark through a valid inequality or an asymptotically finite positive ratio.
1. Topic Focus
Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.
This topic: Transfer convergence or divergence from a known positive benchmark through a valid inequality or an asymptotically finite positive ratio.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Choose a p-series or geometric benchmark from the dominant behavior, then state why the comparison direction supports the conclusion.
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. Comparison tests classify a positive-term series by relating it to a known benchmark, usually a \(p\)-series or geometric series. The comparison must have the correct logical direction.
Direct Comparison Test
Assume \(a_n,b_n\ge0\) for all sufficiently large \(n\).
A smaller positive series is controlled by a convergent ceiling. A larger positive series is forced upward by a divergent floor.
Directions that prove nothing
- Being smaller than a divergent series does not decide convergence.
- Being larger than a convergent series does not decide divergence.
For instance, \(1/n^2<1/n\), but the smaller series converges while the larger one diverges.
Why direct comparison works
Termwise inequalities transfer to partial sums. If \(0\le a_n\le b_n\), then
A bounded increasing sequence of partial sums converges; an increasing sequence forced above an unbounded benchmark diverges.
Build useful inequalities
For positive quantities, a larger denominator produces a smaller reciprocal:
To prove divergence, seek a lower bound such as \(a_n\ge C/n\). Check that all quantities are positive before taking reciprocals or cross-multiplying.
Choose a benchmark from dominant behavior
- Rational powers of \(n\): compare leading powers and use \(1/n^p\).
- Exponentials: compare dominant bases and use a geometric series.
- Bounded factors: use \(|\sin n|\le1\), \(|\cos n|\le1\), or another fixed bound.
- Factorials: a geometric bound may work, though the Ratio Test is often simpler.
Limit Comparison Test
For eventually positive \(a_n,b_n\), if
then \(\sum a_n\) and \(\sum b_n\) have the same convergence behavior. The finite positive limit means \(a_n\) is asymptotically a constant multiple of \(b_n\).
Useful one-sided limit cases
- If \(a_n/b_n\to0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(a_n/b_n\to\infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
The reversed pairings are inconclusive: ratio \(0\) with a divergent benchmark, or ratio \(\infty\) with a convergent benchmark, provides no classification.
Direct or limit comparison?
Use direct comparison when a clean inequality is visible. Use limit comparison when numerator and denominator contain sums whose leading terms clearly determine long-run behavior but a useful global inequality is awkward.
Eventually is enough
Both tests require the comparison only for all sufficiently large \(n\). Finitely many exceptions do not affect convergence, so state a threshold \(n\ge N\) when needed.
Series with signs
Direct and limit comparison are designed for nonnegative terms. For a sign-changing series, they may be applied to \(\sum|a_n|\) to prove absolute convergence. They do not by themselves establish conditional convergence.
Comparison does not find the sum
Even when \(\lim a_n/b_n=1\), the series do not generally have equal sums. The conclusion concerns convergence behavior only.
Comparison checklist
- Apply the nth term test first.
- Confirm eventual nonnegativity.
- Identify the dominant form and choose a known benchmark.
- For direct comparison, write the termwise inequality and check its logical direction.
- For limit comparison, compute the ratio limit and verify its useful case.
- Name the benchmark's convergence behavior and state the transferred conclusion.
6. Detailed Worked Example and Error Check
Example 1: Direct comparison with a convergent \(p\)-series
For positive \(n\),
Since \(\sum1/n^3\) converges, \(\sum1/(n^3+3n+1)\) converges by direct comparison.
Example 2: Direct comparison with a geometric series
The geometric benchmark converges, so \(\sum1/(2^n+1)\) converges.
Example 3: A divergent lower bound
For \(n\ge2\), \(\ln n<n\), hence
The harmonic series diverges, so \(\sum_{n=2}^{\infty}1/\ln n\) diverges by direct comparison.
Example 4: Create a harmonic lower bound
For \(n\ge1\), \(n^2+1\le2n^2\), so
Since \(\sum1/(2n)\) diverges, \(\sum n/(n^2+1)\) diverges.
Example 5: Limit comparison with \(1/n^2\)
Let \(a_n=(3n+1)/(n^3+2)\) and \(b_n=1/n^2\). Then
Since \(0<3<\infty\) and \(\sum b_n\) converges, \(\sum a_n\) converges.
Example 6: Limit comparison with the harmonic series
For \(a_n=(n+2)/(n^2+1)\) and \(b_n=1/n\),
The harmonic benchmark diverges, so the original series diverges.
Example 7: A radical denominator
Compare \(a_n=1/(\sqrt n+1)\) with \(b_n=1/\sqrt n\):
Since the \(p=1/2\) benchmark diverges, the original series diverges.
Example 8: Exponential direct comparison
Because \(3^n+n>3^n\),
The geometric benchmark converges, so the original series converges.
Example 9: Bounded oscillation and absolute convergence
Since \(|\cos n|\le1\),
Thus \(\sum|\cos n|/n^2\) converges, so \(\sum\cos n/n^2\) converges absolutely.
Example 10: An inconclusive comparison choice
If \(a_n=1/n^2\) and the chosen benchmark is \(b_n=1/n^3\), then \(a_n/b_n=n\to\infty\). Because \(\sum b_n\) converges, this limit pairing gives no information. Choosing the exact \(p=2\) form classifies \(\sum a_n\) immediately.
Common errors
- Using a convergent lower bound to claim convergence.
- Using a divergent upper bound to claim divergence.
- Reversing a reciprocal inequality.
- Comparing terms that are not eventually nonnegative.
- Choosing a benchmark from lower-order rather than dominant terms.
- Claiming equal sums from a limit ratio of \(1\).
- Using a ratio limit of \(0\) or infinity without checking the benchmark direction.
- Forcing direct comparison when limit comparison is much cleaner.
- Ignoring that a comparison only needs to hold eventually.
- Using comparison on signed terms without explaining absolute values.
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 a direct or limit comparison and justify the benchmark.
(a) \(\sum_{n=1}^{\infty}1/(n^4+7)\).
(b) \(\sum_{n=1}^{\infty}1/(\sqrt n+5)\).
(c) \(\sum_{n=2}^{\infty}(2n^2+1)/(n^4-3)\).
(d) \(\sum_{n=1}^{\infty}(n^3+1)/(n^4+n)\).
(e) \(\sum_{n=1}^{\infty}4^n/(5^n+n^2)\).
(f) \(\sum_{n=3}^{\infty}1/(\ln n)^2\).
(g) \(\sum_{n=1}^{\infty}\sin^2n/n^2\).
(h) \(\sum_{n=1}^{\infty}(-1)^n/(n^2+1)\).
(i) \(\sum_{n=2}^{\infty}1/(n^2-1)\).
(j) Explain why \(1/(n^2-1)>1/n^2\) does not prove convergence by direct comparison, then name a suitable method.
(k) Use the one-sided limit comparison case on \(a_n=1/n^3\) and \(b_n=1/n^2\).
(l) Use the one-sided limit comparison case on \(a_n=1/\sqrt n\) and \(b_n=1/n\).
Check the solution
(a) \(0<1/(n^4+7)<1/n^4\). The \(p=4\) benchmark converges, so the series converges.
(b) Compare with \(1/\sqrt n\). The ratio \(\sqrt n/(\sqrt n+5)\to1\), and the \(p=1/2\) series diverges, so the given series diverges.
(c) Compare with \(1/n^2\): \(\lim[(2n^2+1)/(n^4-3)]/(1/n^2)=2\). The series converges.
(d) Compare with \(1/n\): \(\lim[(n^3+1)/(n^4+n)]/(1/n)=1\). The series diverges.
(e) Since \(5^n+n^2>5^n\), \(0<4^n/(5^n+n^2)<(4/5)^n\). The series converges.
(f) For sufficiently large \(n\), \((\ln n)^2<n\), so \(1/(\ln n)^2>1/n\). The series diverges by direct comparison.
(g) \(0\le\sin^2n/n^2\le1/n^2\), so the series converges.
(h) \(1/(n^2+1)<1/n^2\), so the absolute-value series converges. The original series converges absolutely.
(i) Compare with \(1/n^2\): \(\lim[n^2/(n^2-1)]=1\). The series converges.
(j) A series larger than a convergent benchmark may converge or diverge, so that inequality has the wrong direction. Limit comparison with \(1/n^2\) gives ratio \(1\) and proves convergence.
(k) \(a_n/b_n=1/n\to0\), and \(\sum b_n=\sum1/n^2\) converges. Therefore \(\sum a_n=\sum1/n^3\) converges.
(l) \(a_n/b_n=\sqrt n\to\infty\), and \(\sum b_n=\sum1/n\) diverges. Therefore \(\sum a_n=\sum1/\sqrt n\) diverges.