harmonic series

Harmonic series

$$\sum _{k=1}^{\infty }\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots =\infty$$

The harmonic series diverges, despite its terms $\frac{1}{k}\to 0$.

Proof of divergence

Group the terms into blocks that double in size:

$$1+\underset{\text{1 term}}{\underbrace{\frac{1}{2}}}+\underset{\text{2 terms}}{\underbrace{\frac{1}{3}+\frac{1}{4}}}+\underset{\text{4 terms}}{\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}}+\underset{\text{8 terms}}{\underbrace{\frac{1}{9}+\cdots +\frac{1}{16}}}+\cdots$$

In each block, every term is at least as large as the last (smallest) term in that block:
$\frac{1}{3}+\frac{1}{4}\ge 2\cdot \frac{1}{4}=\frac{1}{2}$
$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\ge 4\cdot \frac{1}{8}=\frac{1}{2}$
$\frac{1}{9}+\cdots +\frac{1}{16}\ge 8\cdot \frac{1}{16}=\frac{1}{2}$

Each block adds at least $\frac{1}{2}$. There are infinitely many blocks, so the sum grows without bound.

By the $n$-th block we’ve reached index $2^n$, and collected $n$ blocks each worth $\ge \frac{1}{2}$, so:

$$s_{2^n}\ge 1+\frac{n}{2}\to \infty$$

Generalization: $\sum \frac{1}{k^c}$ → p-series

Alternating harmonic series converges

$$\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$$

The alternation makes the partial sums bounce above and below the limit, converging by the Leibniz criterion. But the series is not absolutely convergent (the absolute values give back the divergent harmonic series).