A logarithmic borderline: the series Σ 1/(n^p ln n)
A worksheet asks for which real p the series S(p) = Σ_{n=2}^∞ 1/(n^p ln n) converges. The candidate submits three steps. (1) 'The ratio of consecutive terms tends to 1 for every p, so the ratio test gives no information; hence S(p) diverges for every p.' (2) 'Comparing with the p-series Σ 1/n^p, which converges exactly for p > 1, shows that S(p) converges exactly for p > 1.' (3) 'For p = 1 the partial sums are 2.30 at n = 1000 and 2.99 at n = 10⁶; the increments are already tiny, so the series converges, extremely slowly.' The tutor knows that S(p) converges exactly for p > 1 and that p = 1 is the genuine borderline, where the partial sums grow like ln ln n - unbounded, but so slowly that a quick numerical experiment cannot distinguish growth from convergence.
1. Prove the correct convergence set for S(p), giving a genuine argument at the borderline p = 1.
Read the worked answer
Case p > 1: for n ≥ 2, ln n ≥ ln 2 > 0, so 1/(n^p ln n) ≤ 1/(n^p ln 2). Since Σ 1/n^p converges, S(p) converges by the comparison test.
Case p < 1: for large n, ln n ≤ n^{1−p} (true since n^{p−1} ln n → 0), so n^p ln n ≤ n and 1/(n^p ln n) ≥ 1/n. The harmonic series diverges, so S(p) diverges.
Case p = 1: the terms 1/(n ln n) are positive and decrease to 0, and x ↦ 1/(x ln x) is positive and decreasing on [2, ∞), so the integral test applies: ∫₂^N dx/(x ln x) = ln ln N − ln ln 2 → ∞. Hence Σ 1/(n ln n) diverges.
Conclusion: S(p) converges exactly for p > 1. At p = 1 the terms still tend to 0, which shows that the necessary condition for convergence is far from sufficient.
2. Diagnose the logical status of steps (1), (2) and (3): which conclusions are true and which reasons are invalid?
Read the worked answer
Step (1): the observation a_{n+1}/a_n → 1 is correct, but 'ratio test inconclusive' means only that this particular test cannot decide; it says nothing about divergence. The conclusion 'diverges for every p' is false, since p = 2 gives convergence.
Step (2): the conclusion is true, but the stated reason ignores the factor 1/ln n. Comparison with Σ 1/n^p needs an inequality in the correct direction: for p > 1, 1/(n^p ln n) ≤ 1/(n^p ln 2) works; for p ≤ 1 the comparison with Σ 1/n^p fails because 1/(n^p ln n) > 1/n^p, and the divergence at p = 1 requires the integral test. A right answer resting on a wrong comparison is still a wrong solution.
Step (3): slow growth of the partial sums is not evidence of boundedness. The computation is consistent with the truth: ln ln 10⁶ − ln ln 2 ≈ 2.63 + 0.37 = 3.00, so the numbers confirm the divergence, not convergence. Only a proof of boundedness of the partial sums could establish convergence.
3. For which p does the alternating series Σ_{n=2}^∞ (−1)^n/(n^p ln n) converge, and for which p does it converge absolutely?
Read the worked answer
Absolute convergence is exactly convergence of S(p), namely p > 1.
Conditional convergence: let b_n = 1/(n^p ln n). For p ≥ 0 the sequence b_n is eventually decreasing and b_n → 0 (at p = 0, b_n = 1/ln n decreases to 0 for n ≥ 2), so the Leibniz test gives convergence for every p ≥ 0. Combining, the series converges absolutely for p > 1 and conditionally for 0 ≤ p ≤ 1.
Divergence: for p < 0, write b_n = n^{|p|}/ln n, which tends to infinity, so the terms of the alternating series do not tend to 0 and it diverges (indeed every term of the underlying positive series is eventually greater than 1). Summary: absolute for p > 1, conditional for 0 ≤ p ≤ 1, divergent for p < 0.