Mathematics

Mathematics

Problem

For n≥2n\ge 2, consider the series ∑n=2∞(−1)n−1ln⁡nnp,\sum_{n=2}^{\infty} (-1)^{n-1} \frac{\ln n}{n^{p}}, where pp is a positive constant. For which of the following values of pp does the series converge conditionally but not absolutely? a) p=32p=\dfrac{3}{2} b) p=43p=\dfrac{4}{3} c) p=12p=\dfrac12 d) p=2p=2
Intuitive explanation

This problem asks you to distinguish between absolute and conditional convergence by testing two separate series. For absolute convergence, examine ∑ln⁡nnp\sum \frac{\ln n}{n^p} using the integral test or limit comparison—you need pp large enough that the logarithm doesn't dominate. For conditional convergence, apply the alternating series test to the original series, which requires the terms to decrease to zero. The key insight is that logarithmic growth is "slow" compared to polynomial decay, so there's a sweet spot where the alternating series converges but the absolute series diverges.

Intuitive explanation

This problem asks you to distinguish between absolute and conditional convergence by testing two separate series. For absolute convergence, examine ∑ln⁡nnp\sum \frac{\ln n}{n^p} using the integral test or limit comparison—you need pp large enough that the logarithm doesn't dominate. For conditional convergence, apply the alternating series test to the original series, which requires the terms to decrease to zero. The key insight is that logarithmic growth is "slow" compared to polynomial decay, so there's a sweet spot where the alternating series converges but the absolute series diverges.

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