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Infinite Series

Leibniz Test

Definition

Leibniz Test

If the alternating series ∑n=1∞(−1)n−1an\sum_{n=1}^{\infty} (-1)^{n-1} a_n satisfies:

  1. an≥0a_n \geq 0 (n=1,2,…n = 1, 2, \ldots)
  2. an+1≤ana_{n+1} \leq a_n (for sufficiently large nn)
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0

then the series converges.

Note: This is a sufficient condition. Alternating series that satisfy these conditions must converge.

符号说明
SymbolTypePronunciation/ExplanationMeaning in This Article
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms
lim⁡\limMathematical symbolLimitRepresents limit of sequence or function

Formula

Leibniz Test Conditions

Sufficient conditions for the convergence of alternating series ∑n=1∞(−1)n−1an\sum_{n=1}^{\infty} (-1)^{n-1} a_n:

  1. an≥0a_n \geq 0
  2. an+1≤ana_{n+1} \leq a_n (monotonically decreasing)
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0

Examples

Example 1

Determine the convergence of the series ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}.

Solution: This is an alternating series, where an=1na_n = \frac{1}{n}

  1. an>0a_n > 0 ✓
  2. an+1=1n+1<1n=ana_{n+1} = \frac{1}{n+1} < \frac{1}{n} = a_n ✓
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 ✓

The series satisfies the conditions of the Leibniz test, so it converges.

Exercises

Exercise 1

Determine the convergence of the series ∑n=1∞(−1)n+1n1/2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{1/2}}.

Reference Answer (4 个标签)
series convergence conditional convergence alternating series Leibniz test

Solution Approach: First check for absolute convergence. If the series converges absolutely, then the original series converges.

Detailed Steps:

  1. Consider the absolute value series: ∑n=1∞1n1/2\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}
  2. This is a p-series with p=12≤1p = \frac{1}{2} \leq 1, so it diverges.
  3. Since the absolute value series diverges, we need to check further.
  4. The original series is an alternating series with an=1n1/2a_n = \frac{1}{n^{1/2}}.
  5. Check the Leibniz test conditions:
    • an>0a_n > 0 ✓
    • an+1<ana_{n+1} < a_n ✓ (because 1(n+1)1/2<1n1/2\frac{1}{(n+1)^{1/2}} < \frac{1}{n^{1/2}})
    • lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 ✓
  6. The series satisfies the Leibniz test conditions, so it converges.

Answer: The series converges (conditional convergence).

Summary

Symbols Appearing in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
ana_nMathematical symbolGeneral termThe nth term in the series

Chinese-English Glossary

Chinese TermEnglish TermPhoneticExplanation
Leibniz testLeibniz test/ˈlaɪbnɪts test/Method to determine convergence of alternating series
Alternating seriesalternating series/ˈɔːltəneɪtɪŋ ˈsɪəriːz/Series with alternating positive and negative terms
Convergenceconvergence/kənˈvɜːdʒəns/Series partial sums sequence has a finite limit
Conditional convergenceconditional convergence/kənˈdɪʃənəl kənˈvɜːdʒəns/Series converges but absolute value series diverges
Sufficient conditionsufficient condition/səˈfɪʃənt kənˈdɪʃən/Sufficient condition that guarantees series convergence
PreviousIntegral TestLearn the definition, usage, and applications of the integral test. NextAbsolute and Conditional ConvergenceLearn the definitions of absolute and conditional convergence, and their applications.

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