A student argues that the harmonic series ∑ 1/n converges because its terms approach zero as n → ∞. Which response is correct?
AThe student is wrong — the harmonic series diverges despite its terms approaching zero.
BThe student is right — any series whose terms go to zero must converge.
CThe student is partially right — the series converges conditionally but not absolutely.
DThe p-series test does not apply to the harmonic series, so no conclusion is possible.
The harmonic series (p = 1) is the canonical example proving that terms going to zero is necessary but not sufficient for convergence. The partial sums grow like ln(n) — slowly but without bound — so the series diverges. Option B is the precise misconception the p-series concept is designed to demolish.
Question 2 Multiple Choice
Which of the following p-series converges?
A∑ 1/√n (p = 1/2)
B∑ 1/n (p = 1)
C∑ 1/n^(3/2) (p = 3/2)
D∑ 1/n^(0.9) (p = 0.9)
A p-series converges if and only if p > 1. Only option C has p = 3/2 > 1. Options A (p = 0.5), B (p = 1), and D (p = 0.9) all have p ≤ 1 and diverge. The boundary is strict: p = 1 diverges, and even p = 1.001 converges.
Question 3 True / False
The p-series ∑ 1/n² converges to a finite value.
TTrue
FFalse
Answer: True
True. With p = 2 > 1, the p-series test confirms convergence. The exact sum is π²/6 ≈ 1.645, a famous result known as the Basel problem solved by Euler. This is a concrete illustration that p > 1 produces a genuine finite limit.
Question 4 True / False
If the terms of an infinite series approach zero, the series is expected to converge.
TTrue
FFalse
Answer: False
False — this is the central misconception the p-series highlights. The harmonic series ∑ 1/n has terms 1/n → 0, yet it diverges. Terms going to zero is necessary for convergence (the divergence test says: if terms don't go to zero, the series definitely diverges), but it is not sufficient. The rate of decay matters — 1/n decays too slowly.
Question 5 Short Answer
Why is p = 1 exactly the boundary between convergence and divergence for p-series?
Think about your answer, then reveal below.
Model answer: The boundary comes directly from the integral test: the improper integral ∫₁^∞ 1/x^p dx converges when p > 1 (evaluating to 1/(p−1)) and diverges when p ≤ 1. Since 1/x^p is positive and decreasing, the integral test guarantees the series and the integral share the same convergence fate. So the series ∑ 1/n^p converges exactly when its companion integral converges — precisely when p > 1.
The p = 1 boundary is not arbitrary; it is inherited from the exact threshold at which 1/x^p is integrable from 1 to infinity. For p > 1, the function decays fast enough that accumulated area is finite. At p = 1, area grows like ln(x) without bound. The series mirrors this because the integral test directly links them.