AP Calculus AB/BC · Unit 10 · Topic 10.3 · BC Only
The nth Term Test for Divergence
Use the limit of the individual terms as an immediate necessary-condition check before selecting any other series test.
1. Topic Focus
Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.
This topic: Use the limit of the individual terms as an immediate necessary-condition check before selecting any other series test.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A zero term limit is inconclusive: it is required for convergence but does not guarantee it.
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 nth term test is a fast test for divergence. It checks a condition that every convergent series must satisfy before more specialized tests are considered.
Why convergent series must have vanishing terms
Let
For \(n\ge2\), the newest term is the change between consecutive partial sums:
If \(\sum a_n\) converges to \(S\), then both \(S_n\) and \(S_{n-1}\) approach \(S\). Therefore
The nth term test for divergence
This is the contrapositive of the necessary condition above. It includes finite nonzero limits, infinite limits, and oscillatory or otherwise nonexistent limits.
Decision table
- If \(a_n\to c\ne0\), the series diverges.
- If \(a_n\to+\infty\) or \(-\infty\), the series diverges.
- If \(\lim a_n\) does not exist, the series diverges.
- If \(a_n\to0\), the test is inconclusive and another argument is required.
Necessary does not mean sufficient
The implication goes only one way:
Its converse is false. Both \(1/n\) and \(1/n^2\) approach zero, yet \(\sum1/n\) diverges while \(\sum1/n^2\) converges. A zero limit merely keeps convergence possible.
Use the test first
Before comparison, the Integral Test, the Alternating Series Test, or the Ratio Test, compute \(\lim a_n\). A nonzero or nonexistent limit ends the problem immediately and avoids unnecessary work. If the limit is zero, use the structure of the terms to select a stronger test.
Rational expressions
For a rational function of \(n\), compare leading powers:
- Higher degree in the numerator usually gives unbounded terms.
- Equal degrees give the ratio of leading coefficients.
- Lower degree in the numerator gives limit zero, so the nth term test is inconclusive.
Radicals and conjugates
Divide by the dominant power or rationalize a difference of radicals. For example,
so a series with these terms diverges by the nth term test.
Oscillating terms
Factors such as \((-1)^n\), \(\sin(n\pi/2)\), or \(\cos(n\pi)\) can prevent a term limit from existing. Show this with subsequences when useful: if even and odd terms approach different values, the complete sequence has no limit.
Exponential and logarithmic forms
Compare exponential bases or rewrite logarithmic differences. For example,
The nth term test is then inconclusive even though another argument may establish divergence.
Finite initial terms do not matter
The nth term test concerns behavior as \(n\to\infty\). Changing, deleting, or adding finitely many terms cannot repair a nonzero tail limit and cannot change convergence classification.
AP justification language
A complete conclusion names both the limit and the test:
When the limit is zero, state “the nth term test is inconclusive,” not “the series converges.”
Quick workflow
- Identify the full term \(a_n\), including signs and powers.
- Compute \(\lim\limits_{n\to\infty}a_n\).
- If the result is nonzero, infinite, or nonexistent, conclude divergence.
- If the result is zero, inspect the series and choose another valid test.
- State the conclusion with the test's name and its verified condition.
6. Detailed Worked Example and Error Check
Example 1: Equal-degree rational terms
For \(a_n=n/(2n+1)\), divide by \(n\):
Therefore \(\sum n/(2n+1)\) diverges by the nth term test.
Example 2: Quadratic rational terms
For
the nonzero limit proves that \(\sum a_n\) diverges.
Example 3: A radical quotient
Since \(n>0\),
The associated series diverges by the nth term test.
Example 4: Rationalize a radical difference
Because the term limit is nonzero, \(\sum(\sqrt{n^2+n}-n)\) diverges.
Example 5: Periodic oscillation
For \(a_n=\sin(n\pi/2)\), the values repeat \(1,0,-1,0,\ldots\). The term limit does not exist, so \(\sum\sin(n\pi/2)\) diverges.
Example 6: Alternation with nonvanishing magnitude
Let \(a_n=(-1)^n n/(n+1)\). The even terms approach \(1\) and the odd terms approach \(-1\). Thus \(\lim a_n\) does not exist and the series diverges.
Example 7: Zero limit but divergent series
For \(a_n=1/n\), \(\lim a_n=0\), so the nth term test is inconclusive. The harmonic series nevertheless diverges, as shown later by the Integral Test or comparison arguments.
Example 8: Zero limit and convergent series
For \(a_n=1/n^2\), \(\lim a_n=0\), so the nth term test is again inconclusive. This \(p\)-series converges because \(p=2>1\). Examples 7 and 8 show why a zero limit cannot decide the series.
Example 9: A logarithmic difference
For \(a_n=\ln(n+1)-\ln n\),
The nth term test is inconclusive. However, finite partial sums telescope to \(S_N=\ln(N+1)\), so the series diverges by unbounded partial sums.
Example 10: A parameter controls the conclusion
Consider
The term limit is \(p/4\). If \(p\ne0\), the series diverges by the nth term test. If \(p=0\), the term limit is \(0\), so this test alone is inconclusive.
Common errors
- Concluding convergence when the term limit is zero.
- Testing the limit of partial sums when the question asks for the nth term test.
- Ignoring an alternating factor when computing the term limit.
- Calling an infinite limit equal to zero because the denominator contains \(n\).
- Comparing degrees before simplifying nested radicals or exponentials.
- Applying L'Hopital's Rule directly to a discrete sequence without first identifying a related function.
- Using the test on only the absolute value when the signed terms oscillate.
- Writing “diverges because the limit diverges” without identifying which limit is being evaluated.
- Continuing to a longer test after a nonzero term limit already proves divergence.
- Assuming finite initial terms affect the limiting condition.
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.
Apply the nth term test and state exactly what it proves.
(a) \(\sum_{n=1}^{\infty}(5n-2)/(n+7)\).
(b) \(\sum_{n=1}^{\infty}n^2/(2n^2+1)\).
(c) \(\sum_{n=1}^{\infty}(-1)^n\).
(d) \(\sum_{n=1}^{\infty}\cos(n\pi)\).
(e) \(\sum_{n=1}^{\infty}1/\sqrt n\).
(f) \(\sum_{n=1}^{\infty}n/(n^2+1)\).
(g) \(\sum_{n=1}^{\infty}(3^n+2^n)/3^n\).
(h) \(\sum_{n=1}^{\infty}(n+1)^{1/n}\).
(i) \(\sum_{n=1}^{\infty}e^{-n}\).
(j) \(\sum_{n=1}^{\infty}\ln((n+1)/n)\).
(k) For which values of \(c\) does the nth term test immediately prove that \(\sum_{n=2}^{\infty}(cn+2)/(n-1)\) diverges?
(l) Explain why replacing the first five terms of a series cannot change the conclusion obtained from a nonzero nth-term limit.
Check the solution
(a) The term limit is \(5\ne0\), so the series diverges.
(b) The term limit is \(1/2\ne0\), so the series diverges.
(c) The terms alternate between \(-1\) and \(1\); the limit does not exist, so the series diverges.
(d) Since \(\cos(n\pi)=(-1)^n\), the term limit does not exist and the series diverges.
(e) The term limit is \(0\). The nth term test is inconclusive; the series actually diverges by the \(p\)-series test with \(p=1/2\).
(f) The term limit is \(0\), so the nth term test is inconclusive. Another test is required.
(g) The term is \(1+(2/3)^n\to1\ne0\), so the series diverges.
(h) \((n+1)^{1/n}\to1\ne0\), so the series diverges.
(i) \(e^{-n}\to0\), so the nth term test is inconclusive. The series is geometric and converges.
(j) \(\ln((n+1)/n)=\ln(1+1/n)\to0\), so the test is inconclusive. Its partial sums telescope to \(\ln(N+1)\), showing divergence.
(k) The term limit is \(c\). The test proves divergence for every \(c\ne0\); when \(c=0\), it is inconclusive.
(l) A finite change does not affect the tail limit \(\lim\limits_{n\to\infty}a_n\). If that limit is nonzero, the modified series still diverges by the nth term test.