The Harmonic Series Diverges

The partial sums SnS_n follow ln⁡n+γ\ln n + \gamma upward — very slowly, but without end.

Sn=∑k=1n1kS_n=\sum_{k=1}^{n}\frac{1}{k}
Sn=ln⁡n+γ+o(1)S_n=\ln n+\gamma+o(1)

Part of these learning paths

Examples & Exercises

Use Oresme's grouping to show S2m≥1+m2S_{2^m} \ge 1 + \frac{m}{2}, proving the harmonic series diverges.

S2m≥1+m2→∞S_{2^m} \ge 1 + \frac{m}{2} \to \infty
n=1024=210n = 1024 = 2^{10} — growing above the Oresme bound 1+10/2=61 + 10/2 = 6.
Show proof
  1. Group the terms: 13+14>14+14=12\frac{1}{3}+\frac{1}{4} > \frac{1}{4}+\frac{1}{4} = \frac{1}{2}, 15+⋯+18>4⋅18=12\frac{1}{5}+\cdots+\frac{1}{8} > 4\cdot\frac{1}{8} = \frac{1}{2}. In general 12m−1+1+⋯+12m>2m−1⋅12m=12\frac{1}{2^{m-1}+1}+\cdots+\frac{1}{2^m} > 2^{m-1}\cdot\frac{1}{2^m} = \frac{1}{2}.
  2. Hence S2m=1+12+(13+14)+⋯≥1+m⋅12=1+m2S_{2^m} = 1 + \frac{1}{2} + \left(\frac{1}{3}+\frac{1}{4}\right) + \cdots \ge 1 + m\cdot\frac{1}{2} = 1 + \frac{m}{2}.
  3. As m→∞m \to \infty the bound is unbounded, so (Sn)(S_n) is unbounded. An increasing unbounded sequence diverges, so the harmonic series diverges.

Prove the alternating harmonic series ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} converges via the alternating series test. (Its sum ln⁡2\ln 2 is taken as known.)

∑n=1∞(−1)n+1n=1−12+13−⋯\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = 1-\frac{1}{2}+\frac{1}{3}-\cdots
For contrast: without alternating signs SnS_n grows without bound; with them it converges.
Show proof
  1. Alternating series test: if bn↓0b_n \downarrow 0 then ∑n=1∞(−1)n+1bn\sum_{n=1}^{\infty}(-1)^{n+1}b_n converges. Here bn=1nb_n = \frac{1}{n} decreases to 00.
  2. Directly: the even partial sums T2m=(1−12)+(13−14)+⋯T_{2m} = \left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots group into positive pairs, so they increase and are bounded by 11, hence converge. The odd partial sums T2m+1=T2m+12m+1T_{2m+1} = T_{2m}+\frac{1}{2m+1} tend to the same limit.
  3. Hence the alternating harmonic series converges. (That its sum is ln⁡2\ln 2 is not proved here.)

Prove by the integral test that ∑n=1∞1np\sum_{n=1}^{\infty}\frac{1}{n^p} converges if and only if p>1p > 1.

∑n=1N1np≤1+∫1Ndxxp\sum_{n=1}^{N}\frac{1}{n^p} \le 1 + \int_{1}^{N}\frac{dx}{x^p}
The case p=1p = 1: the divergent side of the integral test.
Show proof
  1. f(x)=x−pf(x) = x^{-p} is positive and decreasing. For n≥2n \ge 2, 1np≤∫n−1ndxxp\frac{1}{n^p} \le \int_{n-1}^{n}\frac{dx}{x^p}, so SN≤1+∫1NdxxpS_N \le 1 + \int_{1}^{N}\frac{dx}{x^p}.
  2. If p>1p > 1, ∫1Ndxxp=1−N1−pp−1<1p−1\int_{1}^{N}\frac{dx}{x^p} = \frac{1 - N^{1-p}}{p-1} < \frac{1}{p-1} is bounded, so the increasing partial sums are bounded and converge.
  3. If p≤1p \le 1, 1np≥1n\frac{1}{n^p} \ge \frac{1}{n}, so comparison with the harmonic series gives divergence.

6. Series