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The World of Limits in Advanced Mathematics

The Monotone Convergence Criterion

The monotone convergence criterion is a foundational criterion in the theory of sequence limits. It establishes the relationship between the monotonicity and boundedness of a sequence, providing an important tool for determining convergence.

The Theorem

The monotone convergence criterion

A monotone and bounded sequence must have a limit: if the sequence {xn}\{x_n\} is monotonically increasing and bounded above, or monotonically decreasing and bounded below, then it has a limit.

Geometric meaning: a monotone and bounded sequence gradually approaches a limit value, which is the supremum or infimum of the sequence.

Idea of the Proof

Take the case of a monotonically increasing sequence bounded above:

  1. Since the sequence is bounded above, by the completeness axiom the sequence has a supremum M=sup⁡{xn}M = \sup\{x_n\}

  2. For any ε>0\varepsilon > 0, there exists NN such that xN>M−εx_N > M - \varepsilon

  3. Since the sequence is monotonically increasing, when n>Nn > N, xn≥xN>M−εx_n \geq x_N > M - \varepsilon

  4. Since MM is the supremum, xn≤Mx_n \leq M

  5. Therefore when n>Nn > N, M−ε<xn≤M<M+εM - \varepsilon < x_n \leq M < M + \varepsilon, i.e., ∣xn−M∣<ε\vert x_n - M \vert < \varepsilon

  6. Hence lim⁡xn=M\lim x_n = M

Application Scenarios

  • The sequence has obvious monotonicity
  • The sequence is bounded
  • The limit cannot be found directly
  • The sequence is defined by a recurrence relation

Practice Problems

Exercise 1

Prove that the sequence xn=n2+1n2+nx_n = \frac{n^2 + 1}{n^2 + n} converges and find its limit.

Reference Answer(1 个标签)
monotone convergence criterion

Idea: First prove that the sequence is monotonically decreasing and bounded below, then find the limit.

Detailed steps:

  1. Prove monotone decreasing: xn+1−xn=(n+1)2+1(n+1)2+(n+1)−n2+1n2+n<0x_{n+1} - x_n = \frac{(n+1)^2 + 1}{(n+1)^2 + (n+1)} - \frac{n^2 + 1}{n^2 + n} < 0

  2. Prove bounded below: xn=n2+1n2+n=1−n−1n2+n>0x_n = \frac{n^2 + 1}{n^2 + n} = 1 - \frac{n-1}{n^2 + n} > 0

  3. By the monotone convergence criterion, the sequence converges

  4. Find the limit: lim⁡n→∞n2+1n2+n=lim⁡n→∞1+1n21+1n=1\lim_{n \to \infty} \frac{n^2 + 1}{n^2 + n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n}} = 1

Answer: The sequence converges, and its limit is 1.

Exercise 2

Find the limit of the sequence xn=nn+1x_n = \frac{n}{n+1}.

Reference Answer(1 个标签)
monotone convergence criterion

Idea: Prove that the sequence is monotonically increasing and bounded above, then find the limit.

Detailed steps:

  1. Prove monotone increasing: xn+1−xn=n+1n+2−nn+1=1(n+1)(n+2)>0x_{n+1} - x_n = \frac{n+1}{n+2} - \frac{n}{n+1} = \frac{1}{(n+1)(n+2)} > 0

  2. Prove bounded above: xn=nn+1=1−1n+1<1x_n = \frac{n}{n+1} = 1 - \frac{1}{n+1} < 1

  3. By the monotone convergence criterion, the sequence converges

  4. Find the limit: lim⁡n→∞nn+1=1\lim_{n \to \infty} \frac{n}{n+1} = 1

Answer: The limit is 1.

Exercise 3

Prove that the sequence xn=(1+1n)nx_n = \left(1 + \frac{1}{n}\right)^n converges.

Reference Answer(1 个标签)
monotone convergence criterion

Idea: Prove that the sequence is monotonically increasing and bounded above.

Detailed steps:

  1. Prove monotone increasing: Expanding by the binomial theorem, one can show that xn+1>xnx_{n+1} > x_n

  2. Prove bounded above: One can show that xn<3x_n < 3 (by mathematical induction)

  3. By the monotone convergence criterion, the sequence converges

  4. The limit of this sequence is the natural constant ee

Answer: The sequence converges, and its limit is ee.

Exercise 4

Prove that the sequence xn=nnx_n = \sqrt[n]{n} converges and find its limit.

Reference Answer(1 个标签)
monotone convergence criterion

Idea: Prove that the sequence is monotonically decreasing and bounded below.

Detailed steps:

  1. Prove monotone decreasing: When n≥3n \geq 3, xn+1<xnx_{n+1} < x_n

  2. Prove bounded below: xn=nn≥1x_n = \sqrt[n]{n} \geq 1

  3. By the monotone convergence criterion, the sequence converges

  4. Find the limit: lim⁡n→∞nn=1\lim_{n \to \infty} \sqrt[n]{n} = 1

Answer: The sequence converges, and its limit is 1.

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
MM数学符号supremumThe smallest upper bound of the sequence
{xn}\{x_n\}数学符号sequenceA sequence
sup⁡\sup数学符号supremumThe supremum of a set
lim⁡\lim数学符号limitThe limit of a function or sequence

中英对照

中文术语英文术语音标说明
单调有界准则monotone bounded theorem/ˈmɒnətəʊn ˈbaʊndɪd ˈθɪərəm/A criterion for determining the convergence of a sequence
单调递增monotone increasing/ˈmɒnətəʊn ɪnˈkriːsɪŋ/Sequence or function values gradually increase
单调递减monotone decreasing/ˈmɒnətəʊn dɪˈkriːsɪŋ/Sequence or function values gradually decrease
有上界bounded above/ˈbaʊndɪd əˈbʌv/There exists a number greater than or equal to all terms
有下界bounded below/ˈbaʊndɪd bɪˈləʊ/There exists a number less than or equal to all terms
上确界supremum/suːˈpriːməm/The smallest upper bound
下确界infimum/ɪnˈfaɪməm/The largest lower bound
确界原理completeness axiom/kəmˈpliːtnəs ˈæksɪəm/A fundamental property of the real number system
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