10.5 Harmonic Series and p-Series

This whole unit is BC only. Σ (n = 1 to ∞) 1/nᵖ converges if p > 1 and diverges if p ≤ 1.

This follows from the integral test: ∫ from 1 to ∞ of 1/xᵖ dx converges exactly when p > 1 (6.13). For p ≤ 0 the terms don't even go to 0, so the nth term test also shows divergence.

The harmonic series Σ (n = 1 to ∞) 1/n diverges, but very slowly. Its partial sums grow roughly like ln n. Adding the first million terms gives only about 14.39. That slow growth fools people into thinking it converges.

The alternating harmonic series, 1 − 1/2 + 1/3 − 1/4 + …, does converge (10.7). It converges to ln 2, though you'll mostly just need the fact that it converges.

• Rewrite roots as powers: 1/∛(n²) = 1/n^(2/3), so p = 2/3 and the series diverges.
• Combine powers: n/n⁴ = 1/n³, so p = 3.
• A constant multiple doesn't change convergence: Σ 5/n² converges because Σ 1/n² does.
• If the term isn't exactly 1/nᵖ, like 1/(n² + 3), compare it to a p-series instead (10.6).

Both families are benchmarks, and they're easy to mix up. In a p-series, n is in the base and the exponent is fixed: 1/n². In a geometric series, n is in the exponent: 1/2ⁿ = (1/2)ⁿ. The tests are different too: p-series need p > 1; geometric series need |r| < 1.

Shifting the index doesn't change anything: Σ (n = 1 to ∞) 1/(n + 1)² is just Σ 1/n² without its first term, so it converges too.

Along with geometric series, p-series are your main benchmarks. Most comparison and limit comparison tests (10.6) boil down to “this series behaves like Σ 1/nᵖ for some p.” A good habit: for any series with rational terms, look at the highest powers of n on the top and bottom. Their difference tells you the p to compare with.