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The World of Limits in Advanced Mathematics

The Cauchy Convergence Criterion

The Cauchy convergence criterion is an important tool for determining the convergence of a sequence. It does not depend on the specific value of the limit, but rather determines convergence through the intrinsic properties of the sequence.

The Theorem

The Cauchy convergence criterion

The sequence {xn}\{x_n\} converges if and only if: for any ε>0\varepsilon > 0, there exists a positive integer NN such that when m,n>Nm, n > N, ∣xm−xn∣<ε|x_m - x_n| < \varepsilon.

Geometric meaning: if a sequence converges, then from some term onward, the distance between any two terms of the sequence can be made arbitrarily small.

Geometric Meaning

The geometric meaning of the Cauchy criterion is: if a sequence converges, then from some term onward, the distance between any two terms of the sequence can be made arbitrarily small.

Idea of the Proof

Necessity (a convergent sequence satisfies the Cauchy condition)

  1. Let lim⁡xn=A\lim x_n = A
  2. For any ε>0\varepsilon > 0, there exists NN such that when n>Nn > N, ∣xn−A∣<ε2\vert x_n - A \vert < \frac{\varepsilon}{2}
  3. When m,n>Nm, n > N, ∣xm−xn∣≤∣xm−A∣+∣xn−A∣<ε2+ε2=ε\vert x_m - x_n \vert \leq \vert x_m - A \vert + \vert x_n - A \vert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

Sufficiency (a sequence satisfying the Cauchy condition converges)

  1. First prove that a sequence satisfying the Cauchy condition is bounded
  2. Using boundedness, construct a subsequence
  3. Prove that the subsequence converges
  4. Using the Cauchy condition, prove that the original sequence converges to the same limit

Application Scenarios

  • The monotonicity of the sequence is not obvious
  • Convergence must be proved without being able to find the limit
  • The sequence has a complicated recurrence relation
  • The sequence involves irrational or transcendental numbers

Practice Problems

Exercise 1

Prove that the sequence xn=∑k=1n1k2x_n = \sum_{k=1}^{n} \frac{1}{k^2} converges.

Reference Answer(1 个标签)
Cauchy criterion

Idea: Use the Cauchy criterion to prove that the sequence satisfies the Cauchy condition.

Detailed steps:

  1. For any ε>0\varepsilon > 0, take N=⌈1ε⌉N = \left\lceil \frac{1}{\varepsilon} \right\rceil

  2. When m>n>Nm > n > N: ∣xm−xn∣=∑k=n+1m1k2<∑k=n+1m1k(k−1)=1n−1m<1n<ε|x_m - x_n| = \sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} \frac{1}{k(k-1)} = \frac{1}{n} - \frac{1}{m} < \frac{1}{n} < \varepsilon

  3. Therefore the sequence satisfies the Cauchy condition and converges

Answer: The sequence converges.

Exercise 2

Prove that the sequence xn=∑k=1n1kx_n = \sum_{k=1}^{n} \frac{1}{k} diverges.

Reference Answer(1 个标签)
Cauchy criterion

Idea: Use the contrapositive of the Cauchy criterion to prove that the sequence does not satisfy the Cauchy condition.

Detailed steps:

  1. Take ε=12\varepsilon = \frac{1}{2}

  2. For any NN, take n=Nn = N, m=2Nm = 2N

  3. ∣xm−xn∣=∑k=N+12N1k≥N⋅12N=12=ε|x_m - x_n| = \sum_{k=N+1}^{2N} \frac{1}{k} \geq N \cdot \frac{1}{2N} = \frac{1}{2} = \varepsilon

  4. Therefore the sequence does not satisfy the Cauchy condition and diverges

Answer: The sequence diverges.

Exercise 3

Prove that the sequence xn=11+122+132+⋯+1n2x_n = \frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} converges.

Reference Answer(1 个标签)
Cauchy criterion

Idea: Use the Cauchy criterion and the comparison test.

Detailed steps:

  1. For any ε>0\varepsilon > 0, take N=⌈1ε⌉N = \left\lceil \frac{1}{\varepsilon} \right\rceil

  2. When m>n>Nm > n > N: ∣xm−xn∣=∑k=n+1m1k2<∑k=n+1m1k(k−1)=1n−1m<1n<ε|x_m - x_n| = \sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} \frac{1}{k(k-1)} = \frac{1}{n} - \frac{1}{m} < \frac{1}{n} < \varepsilon

  3. Therefore the sequence satisfies the Cauchy condition and converges

Answer: The sequence converges.

Exercise 4

Determine whether the sequence xn=sin⁡nx_n = \sin n converges.

Reference Answer(1 个标签)
Cauchy criterion

Idea: Use the contrapositive of the Cauchy criterion to prove that the sequence does not satisfy the Cauchy condition.

Detailed steps:

  1. Take ε=1\varepsilon = 1

  2. For any NN, one can find m,n>Nm, n > N such that ∣sin⁡m−sin⁡n∣≥1|\sin m - \sin n| \geq 1

  3. This is because the range of sin⁡\sin on [0,2π][0, 2\pi] is [−1,1][-1, 1], and it attains values arbitrarily close to any value in this range

  4. Therefore the sequence does not satisfy the Cauchy condition and diverges

Answer: The sequence diverges.

Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
ε\varepsilon希腊字母Epsilon（伊普西隆）An arbitrarily small positive number
NN数学符号positive integerA sufficiently large positive integer
m,nm, n数学符号positive integersTerm indices of the sequence
{xn}\{x_n\}数学符号sequenceA sequence
∣xm−xn∣\vert x_m - x_n \vert数学符号absolute valueThe distance between two terms of the sequence
lim⁡\lim数学符号limitThe limit of a function or sequence
∑\sum数学符号summationThe summation symbol

中英对照

中文术语英文术语音标说明
柯西收敛准则Cauchy convergence criterion/ˈkoʊʃi kənˈvɜːdʒəns kraɪˈtɪəriən/A criterion for determining the convergence of a sequence
充要条件necessary and sufficient condition/nɪˈsesəri ənd səˈfɪʃənt kənˈdɪʃən/A condition that is both necessary and sufficient
必要性necessity/nɪˈsesɪti/A condition a convergent sequence must satisfy
充分性sufficiency/səˈfɪʃənsi/If the condition holds, the sequence converges
有界性boundedness/ˈbaʊndɪdnəs/The property of being bounded
子数列subsequence/ˈsʌbˌsiːkwəns/A sequence formed by selecting some terms from the original sequence
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