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p-Series

Definition

Definition of p-Series

The series n=11np\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 p1p \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 p1p \neq 1, 1+1xpdx=x1p1p1+\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=limblnb=+\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 p1p \neq 1: 1+1xpdx=x1p1p1+\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=lnx1+=+\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=11n2\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=11n3\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=11n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.

Solution: This is a p-series with p=121p = \frac{1}{2} \leq 1

Therefore, the series diverges.

Exercises

Exercise 1

Determine the convergence of the series n=11n3\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=11n3\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=11n4\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=11n4\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=11n\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=11n=n=11n1/2\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=121p = \frac{1}{2} \leq 1, so the series diverges

Answer: The series diverges.

Exercise 4

Determine the convergence of the series n=11n1.5\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=11n1.5\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=11np\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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