Harmonic and p-Series

Definition

The p-series is:

$$\sum _{n=1}^{\infty }\frac{1}{n^p}=1+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+\dots$$

The special case $p=1$ is the harmonic series:

$$\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$$

Intuition

The harmonic series diverges despite its terms going to zero — the terms decrease too slowly. Grouping them into blocks of doubling size shows each block contributes at least $\frac{1}{2}$. For $p>1$ the terms shrink fast enough that the integral comparison yields a finite bound.

Formal Description

Convergence via integral comparison. The integral test gives

$${\int }_1^{\infty }\frac{1}{x^p}dx<\sum _{n=1}^{\infty }\frac{1}{n^p}<1+{\int }_1^{\infty }\frac{1}{x^p}dx.$$

For $p\ne 1$:

$${\int }_1^{\infty }{x}^{-p}dx=\frac{1}{1-p}\left(\underset{x\to \infty }{\lim }{x}^{1-p}-1\right),$$

which converges when $p>1$ and diverges when $p<1$.

For $p=1$: ${\int }_1^{\infty }\frac{1}{x}dx=\ln x{|}_1^{\infty }=\infty$ (diverges).

Grouping proof of harmonic divergence:

$$\begin{array}{rl}& 1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\cdots +\frac{1}{8})+\cdots \\ \ge & 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty .\end{array}$$

Key Results

$$\sum _{n=1}^{\infty }\frac{1}{n^p}\{\begin{array}{ll}\text{converges}& p>1,\\ \text{diverges}& p\le 1.\end{array}$$

The alternating harmonic series converges (by the alternating series test):

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

Applications

The p-series is a standard benchmark for comparison tests. The harmonic series and its alternating form appear in the Taylor Series of $\ln (1+x)$ evaluated at $x=1$.

Trade-offs

The integral test requires a monotone decreasing positive function; it gives convergence or divergence but not the sum. The boundary case $p=1$ (harmonic) is the canonical example where term decay is necessary but insufficient for convergence.

Links

• Sequences and Series
• Taylor Series — the value $\ln 2$ follows from the Taylor series of $\ln (1+x)$ at $x=1$