Does the p-series ∑ 1/n^p converge for p>1?

Does the p-series ∑ 1/n^p converge for p>1?

• A. Yes; converges for p>1
• B. Yes; converges for all p
• C. Yes; converges only for p≥2
• D. No; diverges for all p

Answer: (A)

The key idea is that a p-series behaves differently depending on whether the exponent p is greater than 1. The standard way to see this is the integral test: compare the series ∑ 1/n^p with the improper integral ∫_1^∞ x^(-p) dx. If p > 1, the antiderivative is x^(1−p)/(1−p), and since 1−p < 0, the term x^(1−p) tends to 0 as x grows, yielding a finite value for the integral. That finite value means the series converges. If p ≤ 1, the integral diverges, so the series diverges. Therefore, for any p > 1, the p-series converges.

The key idea is that a p-series behaves differently depending on whether the exponent p is greater than 1. The standard way to see this is the integral test: compare the series ∑ 1/n^p with the improper integral ∫_1^∞ x^(-p) dx. If p > 1, the antiderivative is x^(1−p)/(1−p), and since 1−p < 0, the term x^(1−p) tends to 0 as x grows, yielding a finite value for the integral. That finite value means the series converges. If p ≤ 1, the integral diverges, so the series diverges. Therefore, for any p > 1, the p-series converges.