Harmonic Series and p-Series

The integral test settles one whole family of series at once. These $p$ -series, together with the geometric series of 10.2, are the reference series of Unit 10: the tests in 10.6 work by measuring other series against them.

$\sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots$

The $p$ -series test

A $p$ -series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$ .

For $p > 0$ , $f(x) = \frac{1}{x^p}$ is positive, continuous and decreasing on $[1, \infty)$ , so the integral test applies:

$p > 1$ : $\int_1^{b} x^{-p}\,dx$ $= \frac{1 - b^{1-p}}{p - 1}$ , and $b^{1-p} \to 0$ , so the integral is $\frac{1}{p - 1}$ : finite. The series converges.
$p = 1$ : $\int_1^{b} \frac{dx}{x}$ $= \ln b \to \infty$ . The series diverges.
$0 < p < 1$ : $\int_1^{b} x^{-p}\,dx$ $= \frac{b^{1-p} - 1}{1 - p} \to \infty$ . The series diverges.
$p \le 0$ : the terms $n^{-p}$ do not approach 0, so the series diverges by the nth term test.

Partial sums of the sum of 1 over n to the p, for n = 1 to 40 and three values of p, one dot per partial sum. p = 1/2, vermillion: the sums climb steeply, 11.27 by n = 40, and grow without bound. p = 1, the harmonic series, blue: the sums climb slowly, 4.28 by n = 40, but still grow without bound. p = 2, emerald: the sums level off under the dashed line at π²/6, about 1.645. The series converges exactly when p is greater than 1. — **Figure 1.** Partial sums for three values of $p$ . LevlPrep original diagram.

The harmonic series

The $p$ -series with $p = 1$ is the harmonic series, $\sum \frac{1}{n}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \cdots$ It is the most important divergent series in the course, because its terms approach 0. The boxes of the integral test show $S_n \ge \int_1^{n+1} \frac{dx}{x}$ $= \ln(n + 1)$ , which grows without bound. The growth is slow: in the figure the blue dots reach only about 4.3 after 40 terms, and it takes 12,367 terms to pass 10. Slow growth is still growth: the partial sums pass every number.

Change every other sign and the result is the alternating harmonic series, $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ , which converges (its sum is $\ln 2$ ). The proof comes in 10.7. Keep the two apart: same sizes of terms, opposite behavior.

Recognizing a $p$ -series

Rewrite with exponent rules until the term is $\frac{1}{n^p}$ , then compare $p$ with 1:

Series	Rewritten	$p$	Result
$\sum \frac{1}{\sqrt{n}}$	$\sum \frac{1}{n^{1/2}}$	$\frac{1}{2}$	diverges
$\sum \frac{\sqrt{n}}{n^2}$	$\sum \frac{1}{n^{3/2}}$	$\frac{3}{2}$	converges
$\sum \frac{1}{n\sqrt[3]{n}}$	$\sum \frac{1}{n^{4/3}}$	$\frac{4}{3}$	converges
$\sum \frac{n}{n^2}$	$\sum \frac{1}{n}$	$1$	diverges (harmonic)
$\sum \frac{7}{n^{1.01}}$	$7\sum \frac{1}{n^{1.01}}$	$1.01$	converges

A constant factor does not change convergence. Neither does the starting index: $\sum_{n=5}^{\infty} \frac{1}{n^2}$ converges because $\sum_{n=1}^{\infty} \frac{1}{n^2}$ does.

Worked example. For which $k > 0$ does $\sum_{n=1}^{\infty} \frac{n}{n^{k+2}}$ converge?

Rewrite: $\frac{n}{n^{k+2}}$ $= \frac{1}{n^{k+1}}$ , a $p$ -series with $p = k + 1$ .

It converges when $k + 1 > 1$ , that is $k > 0$ : for every positive $k$ . At $k = 0$ it would be the harmonic series, which diverges.

Combining reference series

The rules from 10.1 let you build on these results. A sum of two convergent series converges; a convergent series plus a divergent one diverges.

Worked example 2. Does $\sum_{n=1}^{\infty} \left(\frac{2}{n^2} + \frac{3}{4^n}\right)$ converge?

First part: $2\sum \frac{1}{n^2}$ , a $p$ -series with $p = 2 > 1$ , converges.

Second part: $\sum \frac{3}{4^n}$ is geometric with $r = \frac{1}{4}$ , $|r| < 1$ , so it converges.

The sum of two convergent series converges, so the whole series converges. Replace the first part by $\frac{2}{n}$ and the answer flips: a divergent part plus a convergent part diverges.

Saying it right

Name the series and its $p$ : " $\sum \frac{1}{n^{3/2}}$ is a $p$ -series with $p = \frac{3}{2} > 1$ , so it converges." Exam readers took points from answers that called a series "harmonic" when it was not, or gave the result with no value of $p$ . And as with the integral test, the test gives convergence, not the sum: $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ is a separate fact, not something the test produces.