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Infinite Series

Leibniz Test

Definition

Leibniz test

For an alternating series ∑n=1∞(−1)n−1an\sum_{n=1}^{\infty} (-1)^{n-1} a_n, if

  1. an≥0a_n \ge 0,
  2. an+1≤ana_{n+1} \le a_n for large nn (monotone decreasing),
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0,

then the series converges (sufficient condition).
Leibniz conditions

Alternating series ∑n=1∞(−1)n−1an\sum_{n=1}^{\infty} (-1)^{n-1} a_n converges if:

  1. an≥0a_n \ge 0
  2. an+1≤ana_{n+1} \le a_n
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0

Example

Example 1

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

Solution: an=1na_n = \frac{1}{n} satisfies an>0a_n>0, an+1<ana_{n+1}<a_n, an→0a_n \to 0; thus it converges.

练习题

练习 1

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

参考答案

思路：Absolute series diverges (p=12p=\frac{1}{2}). Check Leibniz conditions.

步骤：

  1. an=1n1/2>0a_n = \frac{1}{n^{1/2}} > 0
  2. an+1<ana_{n+1} < a_n
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0

All hold ⇒ conditional convergence.

答案：收敛（条件收敛）。

总结

本文出现的符号

符号类型读音/说明在本文中的含义
∑\sum希腊字母Sigma（西格玛）求和符号，表示级数
∞\infty数学符号无穷大表示无穷级数，项数无限
lim⁡\lim数学符号极限表示数列或函数的极限
ana_n数学符号通项级数中第 nn 项

中英对照

中文术语英文术语音标说明
莱布尼茨判别法Leibniz test/ˈlaɪbnɪts test/判断交错级数收敛性的方法
交错级数alternating series/ˈɔːltəneɪtɪŋ ˈsɪəriːz/正负项交替出现的级数
收敛convergence/kənˈvɜːdʒəns/级数部分和序列有有限极限
条件收敛conditional convergence/kənˈdɪʃənəl kənˈvɜːdʒəns/级数收敛但绝对值级数发散的情况
充分条件sufficient condition/səˈfɪʃənt kənˈdɪʃən/保证级数收敛的充分条件
PreviousIntegral TestLearn the integral test, when to apply it, and examples. NextAbsolute vs. Conditional ConvergenceDefinitions, tests, and examples for absolute and conditional convergence.

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