AP Calculus AB/BC · Unit 10 · Topic 10.5 · BC Only
Harmonic Series and p-Series
Recognize reciprocal-power benchmarks, identify their effective exponent, and use the critical value p=1 to classify convergence.
1. Topic Focus
Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.
This topic: Recognize reciprocal-power benchmarks, identify their effective exponent, and use the critical value p=1 to classify convergence.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Rewrite roots and quotients as one power of n before comparing the exponent with 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. Harmonic and \(p\)-series are standard benchmarks. Recognizing them quickly prevents unnecessary use of more complicated convergence tests and prepares the comparisons used later in Unit 10.
The harmonic series
Its terms approach zero, but its partial sums grow without bound. Therefore the harmonic series diverges. It is the boundary \(p=1\) in the larger \(p\)-series family.
The \(p\)-series classification
The inequality is strict. Even an exponent just above \(1\) gives convergence, while \(p=1\) still diverges.
Why \(p=1\) is the threshold
Apply the Integral Test to \(f(x)=x^{-p}\). If \(p\ne1\),
- If \(p>1\), then \(b^{1-p}\to0\), producing a finite limit.
- If \(p<1\), then \(b^{1-p}\to\infty\), so the integral diverges.
- If \(p=1\), then \(\int_1^b dx/x=\ln b\to\infty\).
Thus the integral and the corresponding \(p\)-series share exactly the stated classification.
A second proof that the harmonic series diverges
Group the positive terms in blocks whose lengths double:
Every block after the first contributes at least \(1/2\). Consequently,
which becomes arbitrarily large.
Slow divergence is still divergence
The harmonic partial sums grow roughly like \(\ln N\). Integral bounds give
A long numerical table may look nearly stable, but no finite horizontal limit exists.
Reveal the effective exponent
Rewrite roots and quotients as one power of \(n\):
The exponent belongs to the complete simplified term, not merely to the most visible power in the denominator.
Nonpositive exponents
If \(p=0\), every term equals \(1\). If \(p<0\), then \(1/n^p=n^{-p}\) grows rather than decays. In both cases the terms fail the nth term test, so the series diverges.
Constant multiples and finite starting changes
For \(C\ne0\), multiplying by \(C\) does not change convergence:
Beginning at \(n=5\) instead of \(n=1\) removes only finitely many terms, so the classification also remains unchanged.
Shifted powers are not literally \(p\)-series
A series such as \(\sum1/(n+3)^2\) is not in the exact form \(\sum1/n^p\), but reindexing removes finitely many initial reciprocal squares. It therefore has the same convergence behavior. More complicated expressions that merely resemble \(1/n^p\) require a comparison argument.
Harmonic versus alternating harmonic
The alternating harmonic series is
It converges because of sign cancellation, even though the positive harmonic series diverges. It is not a positive \(p\)-series; its convergence is justified by the Alternating Series Test in Topic 10.7.
Remainder for a convergent \(p\)-series
When \(p>1\), the Integral Test remainder estimate gives
This explains why a \(p\)-series with \(p\) only slightly above \(1\) converges very slowly.
Recognition checklist
- Perform the nth term test mentally.
- Rewrite all radicals and quotients using exponent laws.
- Identify the effective power \(p\).
- Apply the strict threshold \(p>1\).
- Check whether signs, shifts, or extra factors mean another test is actually required.
- State the named benchmark and its verified exponent.
6. Detailed Worked Example and Error Check
Example 1: The harmonic boundary
The series \(\sum1/n\) has \(p=1\). Since \(p\le1\), it diverges even though \(1/n\to0\).
Example 2: A convergent fractional power
For \(\sum1/n^{3/2}\), the exponent is \(p=3/2>1\). Therefore the series converges.
Example 3: A divergent fractional power
For \(\sum1/n^{2/3}\), \(p=2/3<1\). The terms approach zero, but the series diverges.
Example 4: A negative exponent
In \(\sum1/n^{-2}=\sum n^2\), the terms grow without bound. The series diverges by the nth term test, consistent with \(p=-2\le1\).
Example 5: Simplify a numerator radical
The effective exponent is \(p=1/2\), so the series diverges.
Example 6: Combine denominator powers
Since \(4/3>1\), the series converges.
Example 7: A constant multiple
The series \(\sum5/n^\pi\) converges because \(p=\pi>1\). The factor \(5\) changes the sum but not the convergence classification.
Example 8: A later starting index
The series \(\sum_{n=4}^{\infty}1/n^2\) converges because it is the \(p=2\) series with only its first three terms removed.
Example 9: A shifted denominator
For \(\sum_{n=1}^{\infty}1/(n+2)^2\), let \(k=n+2\). Then
a tail of the convergent \(p=2\) series.
Example 10: Bound the tail of a \(p\)-series
For \(\sum1/n^2\), after \(N=10\),
The exact remainder is unknown here, but the integral bounds guarantee its size.
Common errors
- Claiming every reciprocal series converges.
- Using \(a_n\to0\) as proof of convergence.
- Including \(p=1\) in the convergent case.
- Reading \(p\) before simplifying radicals and powers.
- Forgetting that a numerator power reduces the effective denominator exponent.
- Treating a negative \(p\) as a small positive exponent.
- Calling an alternating harmonic series a positive \(p\)-series.
- Assuming a shifted or rational expression is literally a \(p\)-series without reindexing or comparison.
- Believing that slow partial-sum growth implies convergence.
- Using the \(p\)-series test to find an exact sum; it only classifies convergence.
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.
Rewrite each term as needed and classify the series.
(a) \(\sum_{n=1}^{\infty}n^{-5/4}\).
(b) \(\sum_{n=1}^{\infty}1/n\).
(c) \(\sum_{n=1}^{\infty}\sqrt n/n^2\).
(d) \(\sum_{n=1}^{\infty}n^3/\sqrt{n^8}\).
(e) \(\sum_{n=1}^{\infty}1/(n^2\sqrt n)\).
(f) \(\sum_{n=1}^{\infty}\sqrt[3]n/n\).
(g) \(\sum_{n=1}^{\infty}7/n^{1.01}\).
(h) \(\sum_{n=1}^{\infty}1/n^0\).
(i) \(\sum_{n=10}^{\infty}1/n^3\).
(j) Classify \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\) and explain why the positive \(p\)-series rule alone is not its justification.
(k) Give integral upper and lower bounds for the remainder after \(N=20\) of \(\sum1/n^2\).
(l) Use integral comparison to bound the harmonic partial sum \(H_{100}\) between two logarithmic expressions.
Check the solution
(a) \(p=5/4>1\), so the series converges.
(b) This is the harmonic series with \(p=1\), so it diverges.
(c) \(\sqrt n/n^2=1/n^{3/2}\). Since \(p=3/2>1\), it converges.
(d) Since \(\sqrt{n^8}=n^4\) for positive \(n\), the term is \(1/n\). The harmonic series diverges.
(e) \(1/(n^2\sqrt n)=1/n^{5/2}\), so it converges.
(f) \(\sqrt[3]n/n=1/n^{2/3}\), so it diverges.
(g) The constant \(7\) does not affect classification, and \(p=1.01>1\), so the series converges.
(h) Every term is \(1\), so the series diverges; equivalently \(p=0\le1\).
(i) Removing the first nine terms does not affect the \(p=3\) classification, so it converges.
(j) The alternating harmonic series converges by the Alternating Series Test. The positive \(p=1\) series of absolute values diverges, so this is conditional convergence.
(k) \(1/21\le R_{20}\le1/20\).
(l) For decreasing \(1/x\), \(\int_1^{101}dx/x\le H_{100}\le1+\int_1^{100}dx/x\). Hence \(\ln101\le H_{100}\le1+\ln100\).