The p-Series

One family

A p-series is a series $\sum \frac{1}{n^p}=1+\frac{1}{2^p}+\frac{1}{3^p}+\dots$Σ 1/nᵖ = 1 + 1/2ᵖ + 1/3ᵖ + …, for a fixed number p. With p = 1 it is the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\dots$1 + 1/2 + 1/3 + …. With p = 2 it is $1+\frac{1}{4}+\frac{1}{9}+\dots$1 + 1/4 + 1/9 + …. With $p=\frac{1}{2}$p = 1/2 it is $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots$1 + 1/√2 + 1/√3 + …. For every p > 0 the terms shrink to 0, so the nth term test decides nothing, and something sharper is needed.

The answer is one condition on p: $\sum \frac{1}{n^p}$Σ 1/nᵖ converges exactly when p > 1.

Terms as strips

Draw each term as a strip one unit wide. The term $\frac{1}{n}$1/n becomes a strip from x = n to x = n + 1, of height $\frac{1}{n}$1/n, standing on the curve $y=\frac{1}{x}$y = 1/x at its left edge.

The curve falls as x grows, so each strip's top is level with the curve at its left edge and above it everywhere else. The strips cover all the area under $y=\frac{1}{x}$y = 1/x from 1 to n + 1, and more. So $1+\frac{1}{2}+\cdots +\frac{1}{n}$1 + 1/2 + … + 1/n is more than that area.

The terms 1, $\frac{1}{2}$1/2, $\frac{1}{3}$1/3, $\frac{1}{4}$1/4, $\frac{1}{5}$1/5 as strips from x = 1 to x = 6, with the vertical scale stretched. Each touches $y=\frac{1}{x}$y = 1/x at its left edge and stands above it to the right.

Strips below the curve

Move each strip to stand on the curve at its right edge instead, and the strips sit under it. For p = 2, the strips of height $\frac{1}{2^2}$1/2², $\frac{1}{3^2}$1/3², …, $\frac{1}{N^2}$1/N² lie under $y=\frac{1}{x^2}$y = 1/x² from 1 to N.

The area under $y=\frac{1}{x^2}$y = 1/x² from 1 to b is $1-\frac{1}{b}$1 − 1/b: the integral of $x^{-2}$x⁻² is $-x^{-1}$−x⁻¹, and $-\frac{1}{b}-\left(-1\right)=1-\frac{1}{b}$−1/b − (−1) = 1 − 1/b. That is less than 1 for every b. So $\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{N^2}<1$1/2² + 1/3² + … + 1/N² < 1, and adding the first term, $\sum \frac{1}{n^2}$Σ 1/n² up to N is less than 2, for every N.

The partial sums of $\sum \frac{1}{n^2}$Σ 1/n² climb with every term, and they never reach 2. A total that climbs but stays under a ceiling settles on a limit, so $\sum \frac{1}{n^2}$Σ 1/n² converges. Its sum is $\frac{{\pi }^2}{6}$π²/6, about 1.645.

Both pictures say the same thing: for terms that are positive and decreasing, the sum and the area under the curve are finite together or infinite together. That is the integral test.

The terms $\frac{1}{4}$1/4, $\frac{1}{9}$1/9, $\frac{1}{16}$1/16, $\frac{1}{25}$1/25, $\frac{1}{36}$1/36 as strips from x = 1 to x = 6, each touching $y=\frac{1}{x^2}$y = 1/x² at its right edge. All five lie under the curve, whose area from 1 to 6 is $1-\frac{1}{6}=\frac{5}{6}$1 − 1/6 = 5/6.

Where p decides

For p other than 1, the integral of $x^{-p}$x⁻ᵖ is $\frac{x^{1-p}}{1-p}$x^(1 − p)/(1 − p), so the area under $y=\frac{1}{x^p}$y = 1/xᵖ from 1 to b is $\frac{b^{1-p}-1}{1-p}$(b^(1 − p) − 1)/(1 − p).

If p > 1, then 1 − p is negative, and $b^{1-p}=\frac{1}{b^{p-1}}$b^(1 − p) = 1/b^(p − 1) shrinks toward 0 as b grows. The area stays below $\frac{1}{p-1}$1/(p − 1) however far b goes, the sum stays below $1+\frac{1}{p-1}$1 + 1/(p − 1), and the series converges. For p = 2 the bound is 1 + 1 = 2.

If p < 1, then 1 − p is positive and $b^{1-p}$b^(1 − p) grows without bound, so the area does, and so does the sum, which is larger than the area. The series diverges.

At p = 1 this formula divides by 0. That case is the harmonic series, and the grouping argument in The nth Term Test for Divergence already shows it diverges. The boundary belongs to the divergent side.

What the partial sums show

For p = 2 the partial sums after 10, 100, 1000 and 10000 terms are about 1.550, 1.635, 1.644 and 1.645: settled. For p = 1 they are about 2.93, 5.19, 7.49 and 9.79: still climbing, by about 2.3 each time the number of terms is multiplied by 10. For $p=\frac{1}{2}$p = 1/2 they are about 5.0, 18.6, 61.8 and 198.5.

Numbers like these illustrate; they do not prove. For p = 1.1 the series converges, by the bound, to less than $1+\frac{1}{0.1}=11$1 + 1/0.1 = 11, yet after 10000 terms its partial sum is only about 6.60 and still rising. No list of partial sums can tell slow convergence from slow divergence; the comparison with the area can.

The usual mistakes

Including p = 1. The condition is p > 1, not $p\ge 1$p ≥ 1: at p = 1 the series is the harmonic series, which diverges.

Taking p > 0 as enough. For $p=\frac{1}{2}$p = 1/2 the terms shrink to 0 and the series diverges.

Confusing $\sum \frac{1}{n^p}$Σ 1/nᵖ with $\sum \frac{1}{p^n}$Σ 1/pⁿ. $\sum \frac{1}{2^n}$Σ 1/2ⁿ is a geometric series with ratio $\frac{1}{2}$1/2 and converges; $\sum \frac{1}{n^2}$Σ 1/n² is a p-series. The power is on n in one and on the constant in the other.

Judging convergence from a few partial sums. After 10000 terms the convergent p = 1.1 series is smaller than the divergent harmonic series.

Planks over the edge of a bench

In the application below, the overhang of a stack of planks is 15 times a partial sum of the harmonic series. Because p = 1, the overhang has no limit, and adding terms finds how many planks reach 1 meter.