p-Series

Definition

Definition of p-Series

The series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} is called a p-series, where pp is a real number.

符号说明
SymbolTypePronunciation/ExplanationMeaning in This Article
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms
ppMathematical symbolParameterParameter of p-series, determining convergence

Convergence

p-Series Convergence

The series converges when p>1p > 1 and diverges when p≤1p \leq 1.

证明
  1. Let f(x)=1xpf(x) = \frac{1}{x^p}, which is continuous, monotonically decreasing, and non-negative on [1,+∞)[1,+\infty).
  2. Compute the integral: when p≠1p \neq 1, ∫1+∞1xpdx=x1−p1−p∣1+∞\int_1^{+\infty} \frac{1}{x^p} dx = \frac{x^{1-p}}{1-p} \Big|_1^{+\infty}
  3. When p>1p > 1, the integral converges, implying the series converges; when p<1p < 1, the integral diverges, so the series diverges.
  4. When p=1p = 1, ∫1+∞1xdx=lim⁡b→∞ln⁡b=+∞\int_1^{+\infty} \frac{1}{x} dx = \lim_{b\to\infty} \ln b = +\infty so the series diverges.

Proof

Using the integral test:

Let f(x)=1xpf(x) = \frac{1}{x^p}, then f(x)f(x) is continuous, monotonically decreasing, and non-negative on [1,+∞)[1, +\infty).

The convergence of the integral ∫1+∞1xpdx\int_1^{+\infty} \frac{1}{x^p} dx:

When p≠1p \neq 1: ∫1+∞1xpdx=x1−p1−p∣1+∞\int_1^{+\infty} \frac{1}{x^p} dx = \frac{x^{1-p}}{1-p} \big|_1^{+\infty}

  • When p>1p > 1, the integral converges
  • When p<1p < 1, the integral diverges

When p=1p = 1: ∫1+∞1xdx=ln⁡x∣1+∞=+∞\int_1^{+\infty} \frac{1}{x} dx = \ln x \big|_1^{+\infty} = +\infty

So the integral diverges.

Examples

Example 1: Determine the convergence of the series ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}.

Solution: This is a p-series with p=2>1p = 2 > 1

Therefore, the series converges.

Example 2: Determine the convergence of the series ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3}.

Solution: This is a p-series with p=3>1p = 3 > 1

Therefore, the series converges.

Example 3: Determine the convergence of the series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.

Solution: This is a p-series with p=12≤1p = \frac{1}{2} \leq 1

Therefore, the series diverges.

Exercises

Exercise 1

Determine the convergence of the series ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3}.

Reference Answer (2 个标签)
p-series series convergence

Problem-solving approach: This is a p-series; we need to determine the relationship between p and 1.

Detailed steps:

  1. Identify the series type: ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3} is a p-series
  2. Determine the p value: p=3p = 3
  3. Check convergence: p=3>1p = 3 > 1, so the series converges

Answer: The series converges.

Exercise 2

Determine the convergence of the series ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^4}.

Reference Answer (2 个标签)
p-series series convergence

Problem-solving approach: This is a p-series; we need to determine the relationship between p and 1.

Detailed steps:

  1. Identify the series type: ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^4} is a p-series
  2. Determine the p value: p=4p = 4
  3. Check convergence: p=4>1p = 4 > 1, so the series converges

Answer: The series converges.

Exercise 3

Determine the convergence of the series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.

Reference Answer (2 个标签)
p-series series convergence

Problem-solving approach: This is a p-series; we need to determine the relationship between p and 1.

Detailed steps:

  1. Identify the series type: ∑n=1∞1n=∑n=1∞1n1/2\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} is a p-series
  2. Determine the p value: p=12p = \frac{1}{2}
  3. Check convergence: p=12≤1p = \frac{1}{2} \leq 1, so the series diverges

Answer: The series diverges.

Exercise 4

Determine the convergence of the series ∑n=1∞1n1.5\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}.

Reference Answer (2 个标签)
p-series series convergence

Problem-solving approach: This is a p-series; we need to determine the relationship between p and 1.

Detailed steps:

  1. Identify the series type: ∑n=1∞1n1.5\sum_{n=1}^{\infty} \frac{1}{n^{1.5}} is a p-series
  2. Determine the p value: p=1.5p = 1.5
  3. Check convergence: p=1.5>1p = 1.5 > 1, so the series converges

Answer: The series converges.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
nnMathematical symbolNumber of termsNumber of terms in the series
∫\intMathematical symbolIntegralRepresents definite or indefinite integral
ln⁡\lnMathematical symbolNatural logarithmNatural logarithm function
lim⁡\limMathematical symbolLimitRepresents limit of sequence or function

Chinese-English Glossary

Chinese TermEnglish TermIPA PronunciationExplanation
pp 级数pp-series/piː ˈsɪəriːz/Series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}
收敛convergence/kənˈvɜːdʒəns/Partial sums sequence has a finite limit
发散divergence/daɪˈvɜːdʒəns/Partial sums sequence has no finite limit
积分判别法integral test/ˈɪntɪɡrəl test/Method to determine series convergence using integrals
比较判别法comparison test/kəmˈpærɪsən test/Method to determine series convergence by comparison

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