p-Series

A p-series is a series of the form \sum_{n=1}^{\infty} \dfrac{1}{n^p}, where p is a constant. The p-series test says it converges when p > 1 and diverges when p ≤ 1. For example, \sum 1/n^2 converges, while the harmonic series \sum 1/n (p = 1) and \sum 1/\sqrt{n} diverge.

Before this lesson: Integral Test, Divergence Test and Rational Exponents

Why the power p decides everything

Every p-series has terms that shrink toward 0 when p > 0. The question is whether they shrink fast enough for the running total to level off.

Compare three series after 20 terms. The partial sums of \sum 1/\sqrt{n} have reached about 7.6 and keep climbing quickly. The harmonic series \sum 1/n has reached about 3.6 and keeps climbing slowly. The partial sums of \sum 1/n^2 have reached about 1.6 and are leveling off.

The cutoff is exactly p = 1. Terms like 1/n^{1.01} shrink just fast enough, and terms like 1/n^{0.99} do not.

The p-series test

\sum_{n=1}^{\infty} \dfrac{1}{n^p} \text{ converges if } p \gt 1 \text{ and diverges if } p \le 1

Why it works. For p > 0, the function f(x) = x^{-p} is positive, continuous and decreasing, so the integral test applies. For p ≠ 1:

\int_1^{b} x^{-p}\,dx = \dfrac{b^{1-p} - 1}{1 - p}

If p > 1, the exponent 1 − p is negative, so b^{1-p} \to 0 and the integral equals \dfrac{1}{p - 1}, a finite number. If p < 1, b^{1-p} \to \infty, so the integral diverges. For p = 1 the integral is ln b, which also diverges. For p ≤ 0 the terms do not go to 0, so the divergence test shows divergence.

Constants and shifts. A constant multiple, like \sum 5/n^2, behaves the same as the p-series. So does a shifted series like \sum 1/(n + 3)^2, which is the p-series with its first three terms missing.

How to use the p-series test

1. Simplify the term into the form \dfrac{c}{n^p}, using the exponent rules to combine powers of n.
2. Read off p.
3. If p > 1, the series converges. If p ≤ 1, it diverges.

p-series examples

Example 1: A large power

Does \sum_{n=1}^{\infty} \dfrac{1}{n^4} converge?

p = 4

4 \gt 1

Answer: the series converges by the p-series test. Its terms 1, 1/16, 1/81, … shrink very quickly.

Example 2: A square root in the denominator

Does \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} converge?

\dfrac{1}{\sqrt{n}} = \dfrac{1}{n^{1/2}}

p = \dfrac{1}{2}

\dfrac{1}{2} \le 1

Answer: the series diverges, even though its terms go to 0.

Example 3: Combining powers first

Does \sum_{n=1}^{\infty} \dfrac{\sqrt[3]{n}}{n^2} converge?

\dfrac{\sqrt[3]{n}}{n^2} = \dfrac{n^{1/3}}{n^2}

\dfrac{n^{1/3}}{n^2} = n^{1/3 - 2}

n^{1/3 - 2} = n^{-5/3}

n^{-5/3} = \dfrac{1}{n^{5/3}}

p = \dfrac{5}{3} \gt 1

Answer: the series converges by the p-series test.

Example 4: A constant multiple and a shift

Decide whether each series converges: (a) \sum_{n=1}^{\infty} \dfrac{5}{n\sqrt{n}} and (b) \sum_{n=1}^{\infty} \dfrac{1}{(n + 3)^2}.

Part (a)

n\sqrt{n} = n^1 \cdot n^{1/2} = n^{3/2}

\dfrac{5}{n\sqrt{n}} = 5 \cdot \dfrac{1}{n^{3/2}}

p = \dfrac{3}{2} \gt 1

Part (b)

k = n + 3

\sum_{n=1}^{\infty} \dfrac{1}{(n + 3)^2} = \sum_{k=4}^{\infty} \dfrac{1}{k^2}

p = 2 \gt 1

Answer: both series converge.

Example 5: Stacking books (word problem)

You stack identical books, each 1 unit long, at the edge of a table. With the best stacking, the book that is nth from the top can stick out \dfrac{1}{2n} units past the book below it. (a) Can the top book stick out as far as you want past the table? (b) How many books does it take for the top book to stick out more than 2 units?

\sum_{n=1}^{\infty} \dfrac{1}{2n} = \dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{1}{n}

p = 1 \le 1

\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{1}{n} = \infty

\dfrac{1}{2}\left(1 + \tfrac{1}{2} + \cdots + \tfrac{1}{N}\right) \gt 2

1 + \tfrac{1}{2} + \cdots + \tfrac{1}{N} \gt 4

1 + \tfrac{1}{2} + \cdots + \tfrac{1}{30} \approx 3.995

1 + \tfrac{1}{2} + \cdots + \tfrac{1}{31} \approx 4.027

Answer: (a) yes, because the harmonic series diverges, the overhang can be as large as you like. (b) It takes 31 books to get an overhang of more than 2 book lengths. Each extra unit of overhang takes far more books than the one before.

Common mistakes

Related methods

p-series as benchmarks

The main use of p-series is as a yardstick. Many series behave like a p-series for large n. For example, \frac{1}{n^2 + 1} is a little smaller than \frac{1}{n^2}, so its series should converge too. The direct comparison test and the limit comparison test make this idea exact.

Known sums

Only a few p-series have known sums. Two famous ones are \sum 1/n^2 = \pi^2/6 and \sum 1/n^4 = \pi^4/90. For most others, estimate the sum with the remainder estimate from the integral test.

Practice problems

1. Does \sum_{n=1}^{\infty} \dfrac{1}{n^3} converge?
2. Does \sum_{n=1}^{\infty} \dfrac{1}{\sqrt[3]{n}} converge?
3. Does \sum_{n=1}^{\infty} \dfrac{1}{n^{0.9}} converge?
4. Does \sum_{n=1}^{\infty} \dfrac{7}{n^{2.5}} converge?
5. Does \sum_{n=1}^{\infty} \dfrac{\sqrt{n}}{n^2} converge?
6. Does \sum_{n=1}^{\infty} \dfrac{n}{\sqrt[3]{n^4}} converge?
7. Does \sum_{n=1}^{\infty} \dfrac{1}{(n + 2)^3} converge?
8. Does \sum_{n=1}^{\infty} \left(\dfrac{1}{n^2} + \dfrac{1}{2^n}\right) converge?
9. For which values of p does \sum_{n=1}^{\infty} \dfrac{1}{n^{2p - 1}} converge?
10. A student adds the first 10 terms of \sum_{n=1}^{\infty} \dfrac{1}{n^2} and gets about 1.54977. Use the remainder estimate to trap the full sum between two numbers.

p = 3 \gt 1

\dfrac{1}{\sqrt[3]{n}} = \dfrac{1}{n^{1/3}}

p = \dfrac{1}{3} \le 1

p = 0.9 \le 1

\dfrac{7}{n^{2.5}} = 7 \cdot \dfrac{1}{n^{2.5}}

p = 2.5 \gt 1

\dfrac{\sqrt{n}}{n^2} = \dfrac{n^{1/2}}{n^2} = n^{1/2 - 2}

n^{1/2 - 2} = n^{-3/2} = \dfrac{1}{n^{3/2}}

p = \dfrac{3}{2} \gt 1

\dfrac{n}{\sqrt[3]{n^4}} = \dfrac{n^1}{n^{4/3}}

\dfrac{n^1}{n^{4/3}} = n^{1 - 4/3} = n^{-1/3}

n^{-1/3} = \dfrac{1}{n^{1/3}}

p = \dfrac{1}{3} \le 1

\sum_{n=1}^{\infty} \dfrac{1}{(n + 2)^3} = \sum_{k=3}^{\infty} \dfrac{1}{k^3}

p = 3 \gt 1

\sum \dfrac{1}{n^2} \text{ converges}

\sum \dfrac{1}{2^n} \text{ converges}

\sum \left(\dfrac{1}{n^2} + \dfrac{1}{2^n}\right) \text{ converges}

2p - 1 \gt 1

2p \gt 2

p \gt 1

\int_{11}^{\infty} \dfrac{dx}{x^2} \le R_{10} \le \int_{10}^{\infty} \dfrac{dx}{x^2}

\dfrac{1}{11} \le R_{10} \le \dfrac{1}{10}

0.09091 \le R_{10} \le 0.1

1.54977 + 0.09091 \le S \le 1.54977 + 0.1

1.64068 \le S \le 1.64977

p-series FAQ

What is a p-series?

A p-series is a series of the form \sum 1/n^p, where p is a fixed number. Examples are \sum 1/n^2, \sum 1/\sqrt{n} and the harmonic series \sum 1/n.

When does a p-series converge?

It converges when p > 1 and diverges when p ≤ 1. The proof uses the integral test on f(x) = 1/x^p.

Is the harmonic series a p-series?

Yes. The harmonic series \sum 1/n is the p-series with p = 1, and it diverges. It is the boundary case of the test.

What is the difference between a p-series and a geometric series?

In a p-series the variable is the base, as in 1/n^2. In a geometric series the variable is the exponent, as in 1/2^n. A geometric series converges when its ratio has size less than 1.

Summary

• A p-series has the form \sum 1/n^p.
• It converges if p > 1 and diverges if p ≤ 1.
• The harmonic series (p = 1) diverges, even though its terms go to 0.
• Simplify roots and powers into c/n^p before reading off p.
• Constant multiples and shifted versions of a p-series behave the same way.

Previous lesson: Integral Test

Next lesson: Direct Comparison Test

Back to the Calculus 2 Index