Unit 10 · Topic 10.5

Harmonic Series and p-Series

A p -series ∑ 1/(n^p) converges exactly when p > 1 ; the integral test proves it.

Practice 3: JustificationPractice 4: Communication and Notation

Part 1 · Hook

Why this matters

Add 1+12+13+14+⋯1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots and the total creeps up so slowly that after a million terms it is barely past 14, yet it never stops growing. Square each denominator, 1+14+19+⋯1 + \frac{1}{4} + \frac{1}{9} + \cdots, and the total settles near 1.645. One family of series, ∑1np\sum \frac{1}{n^p}, sits on that boundary, and it is the yardstick for comparing every other series.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If ff is positive, continuous and decreasing and ∫1∞f(x)dx\int_1^{\infty} f(x)\,dx diverges, then ∑n=1∞f(n)\sum_{n=1}^{\infty} f(n)

  1. diverges
  2. converges
  3. converges to the integral
  4. may do either
Show the answer

By the integral test, the series and the integral behave alike.

  • Correct: diverges:
  • converges:
  • converges to the integral:
  • may do either:

2. ∫1b1xdx\int_1^{b} \frac{1}{x}\,dx==

  1. ln⁡b\ln b
  2. 1−1b1 - \frac{1}{b}
  3. 1b2\frac{1}{b^2}
  4. ln⁡b−1\ln b - 1
Show the answer

ln⁡b−ln⁡1=ln⁡b\ln b - \ln 1 = \ln b, which grows without bound.

  • Correct: ln⁡b\ln b:
  • 1−1b1 - \frac{1}{b}:
  • 1b2\frac{1}{b^2}:
  • ln⁡b−1\ln b - 1:

Part 4 · See it

See it first

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.
Partial sums of ∑1np\sum \frac{1}{n^p} for p=12p = \frac{1}{2}, 11 and 22: only p=2p = 2 levels off. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Apply the integral test to f(x)=1xpf(x) = \frac{1}{x^p}, p>0p > 0ff is positive, continuous and decreasing on [1,∞)[1, \infty)
  2. For p>1p > 1, ∫1∞x−pdx\int_1^{\infty} x^{-p}\,dx=1p−1= \frac{1}{p - 1} is finitethe pp-series converges
  3. For p=1p = 1, ∫1bdxx\int_1^{b} \frac{dx}{x}=ln⁡b→∞= \ln b \to \infty; for 0<p<10 < p < 1, b1−p−11−p→∞\frac{b^{1-p} - 1}{1 - p} \to \inftythe pp-series diverges
  4. For p≤0p \le 0, the terms do not approach 0the series diverges by the nth term test

Part 6 · Key ideas

Key ideas

  • ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if p>1p > 1 and diverges if p≤1p \le 1.
  • The harmonic series ∑1n\sum \frac{1}{n} (p=1p = 1) diverges, though its terms approach 0. The alternating harmonic series ∑(−1)n+1n\sum \frac{(-1)^{n+1}}{n} converges.
  • Rewrite before deciding: 1n3=1n1/3\frac{1}{\sqrt[3]{n}} = \frac{1}{n^{1/3}}, nn2=1n3/2\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}, n2n3=1n\frac{n^2}{n^3} = \frac{1}{n}. Name the series and its pp when you use it.

Part 7 · Misconception

A common mistake

The wrong idea: ∑1n\sum \frac{1}{n} converges because its terms get smaller and smaller.

What actually happens: The harmonic series diverges: its partial sums pass every number, though very slowly (Sn≥ln⁡(n+1)S_n \ge \ln(n + 1)). Shrinking terms are not enough.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Five series of powers of nn

Five series, each starting at n=1n = 1 (kk is a positive constant)
NameSeries
I∑1n3/2\sum \frac{1}{n^{3/2}}
II∑1n3\sum \frac{1}{\sqrt[3]{n}}
III∑n2n3\sum \frac{n^2}{n^3}
IV∑1nπ\sum \frac{1}{n^{\pi}}
V∑1n2k\sum \frac{1}{n^{2k}}

1. Which of the series I, II, III and IV converge?

  1. I, II and IV
  2. I, III and IV
  3. IV alone
  4. I and IV
Show the answer

A pp-series ∑1np\sum \frac{1}{n^p} converges exactly when p>1p > 1. I: p=32p = \frac{3}{2}. II: p=13p = \frac{1}{3}. III: n2n3=1n\frac{n^2}{n^3} = \frac{1}{n}, p=1p = 1. IV: p=πp = \pi. So I and IV converge.

  • I, II and IV: Series II has p=13p = \frac{1}{3}, which is not greater than 1.
  • I, III and IV: Series III simplifies to 1n\frac{1}{n}: p=1p = 1, the harmonic series, which diverges.
  • IV alone: Series I has p=32>1p = \frac{3}{2} > 1, so it converges too.
  • Correct: I and IV: Right: p=32p = \frac{3}{2} and p=πp = \pi are both greater than 1.

2. What is series III, and does it converge?

  1. A pp-series with p=5p = 5; it converges
  2. A pp-series with p=23p = \frac{2}{3}; it diverges
  3. The harmonic series ∑1n\sum \frac{1}{n}; it diverges
  4. The harmonic series ∑1n\sum \frac{1}{n}; it converges
Show the answer

n2n3=1n\frac{n^2}{n^3} = \frac{1}{n}: the harmonic series, the pp-series with p=1p = 1, which diverges.

  • A pp-series with p=5p = 5; it converges: Simplify first: the powers subtract, n2n3=n−1\frac{n^2}{n^3} = n^{-1}.
  • A pp-series with p=23p = \frac{2}{3}; it diverges: n2n3\frac{n^2}{n^3} is not n−2/3n^{-2/3}: subtract the exponents, 2−3=−12 - 3 = -1.
  • Correct: The harmonic series ∑1n\sum \frac{1}{n}; it diverges: Right: n2n3=1n\frac{n^2}{n^3} = \frac{1}{n}, so p=1p = 1.
  • The harmonic series ∑1n\sum \frac{1}{n}; it converges: The harmonic series diverges, though its terms approach 0.

3. For which values of kk does series V converge? Give an interval.

Type the interval, in interval notation such as [1, 5) or as an inequality such as 1 ≤ x < 5.

Show the answer

Series V is a pp-series with p=2kp = 2k. It converges when 2k>12k > 1, that is k>12k > \frac{1}{2}.

  • Answer: (12,∞)(\frac{1}{2}, \infty)

Graph

Partial sums of the harmonic series

The dots are the partial sums Sn=1+12+⋯+1nS_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} of the harmonic series for n=1n = 1 to 2020. The dashed curve is y=ln⁡(x+1)y = \ln(x + 1); by the integral test picture, Sn≥ln⁡(n+1)S_n \ge \ln(n + 1) for every nn.
0246810121416182000.511.522.533.54nS_ny = ln(x + 1)

4. What does the graph, with Sn≥ln⁡(n+1)S_n \ge \ln(n + 1), show about the harmonic series?

  1. It converges: the partial sums grow more and more slowly
  2. It converges to about 3.6, the last partial sum shown
  3. Nothing: the terms approach 0
  4. It diverges: its lower bound grows without bound
Show the answer

ln⁡(n+1)→∞\ln(n + 1) \to \infty and Sn≥ln⁡(n+1)S_n \ge \ln(n + 1), so Sn→∞S_n \to \infty: the harmonic series diverges.

  • It converges: the partial sums grow more and more slowly: Slowing growth is not leveling off; SnS_n passes every number eventually.
  • It converges to about 3.6, the last partial sum shown: S20≈3.6S_{20} \approx 3.6, but the partial sums keep growing past it.
  • Nothing: the terms approach 0: The nth term test is silent, but the lower bound ln⁡(n+1)\ln(n + 1) settles it.
  • Correct: It diverges: its lower bound grows without bound: Right: a lower bound that grows without bound drags the partial sums up with it.

5. Which statement about the harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} is true?

  1. Its terms approach 0, but its partial sums grow without bound
  2. Its terms approach 0, so it converges to a finite sum
  3. Its partial sums approach ln⁡2\ln 2
  4. It diverges because its terms do not approach 0
Show the answer

1n→0\frac{1}{n} \to 0, yet Sn≥ln⁡(n+1)→∞S_n \ge \ln(n + 1) \to \infty. The harmonic series diverges, which is why the nth term test can never prove convergence.

  • Correct: Its terms approach 0, but its partial sums grow without bound: Right: the standard example that terms approaching 0 do not guarantee convergence.
  • Its terms approach 0, so it converges to a finite sum: Terms approaching 0 are necessary for convergence, not enough.
  • Its partial sums approach ln⁡2\ln 2: ln⁡2\ln 2 is the sum of the alternating harmonic series, not of this one.
  • It diverges because its terms do not approach 0: Its terms 1n\frac{1}{n} do approach 0; it diverges for another reason.

6. How do ∑n=1∞1n0.999\sum_{n=1}^{\infty} \frac{1}{n^{0.999}} and ∑n=1∞1n1.001\sum_{n=1}^{\infty} \frac{1}{n^{1.001}} behave?

  1. Both converge: the powers are close to 1
  2. Both diverge: the powers are close to 1
  3. The first diverges and the second converges
  4. The first converges and the second diverges
Show the answer

A pp-series converges exactly when p>1p > 1: p=0.999p = 0.999 diverges and p=1.001p = 1.001 converges, however close they look.

  • Both converge: the powers are close to 1: How close pp is to 1 does not matter; which side of 1 it is on does.
  • Both diverge: the powers are close to 1: p=1.001>1p = 1.001 > 1, so the second converges, slowly.
  • Correct: The first diverges and the second converges: Right: 0.999≤10.999 \le 1 and 1.001>11.001 > 1.
  • The first converges and the second diverges: Reversed: a larger power makes the terms shrink faster, which helps convergence.

Part 9 · Summary

Summary

A pp-series ∑1np\sum \frac{1}{n^p} converges exactly when p>1p > 1; the integral test proves it. The harmonic series (p=1p = 1) diverges and the alternating harmonic series converges. Simplify a series to the form 1np\frac{1}{n^p}, then compare pp with 1.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic