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Sequences

The Definition of Limits

The definition of a limit has both an intuitive and a rigorous form. Understanding both is important for mastering the concept of limits.

The Intuitive Definition

The intuitive definition of a limit

If, as nn increases without bound, the terms ana_n of the sequence {an}\{a_n\} get infinitely close to a certain constant AA, then the sequence {an}\{a_n\} is said to converge to AA, and AA is called the limit of the sequence.

Written as: lim⁡n→∞an=A\lim_{n \to \infty} a_n = A or an→A(n→∞)a_n \to A \quad (n \to \infty)

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡\limOperatorlimitThe limit operator
∞\inftySymbolinfinityInfinity; n→∞n \to \infty means nn tends to infinity

The Rigorous Definition (the ε-N Definition)

The ε-N definition of a limit

Let {an}\{a_n\} be a sequence and AA a constant. If for any given positive number ε\varepsilon (however small), there always exists a positive integer NN such that whenever n>Nn > N,

∣an−A∣<ε|a_n - A| < \varepsilon

then the sequence {an}\{a_n\} is said to converge to AA, written as lim⁡n→∞an=A\lim_{n \to \infty} a_n = A.

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
ε\varepsilonGreek letterepsilonAn arbitrarily small positive number specifying the degree of closeness
NNVariableNA positive integer indicating from which term the condition holds

The intuitive definition is easy to understand, but not precise enough:

  • What does “infinitely close” mean?
  • To what extent does “increase without bound” go?

The rigorous ε-N definition resolves these issues:

  • ∣an−A∣<ε|a_n - A| < \varepsilon quantifies “closeness”
  • n>Nn > N quantifies “large enough”

This rigor is the foundation of mathematical analysis, allowing us to carry out rigorous proofs.

Analogy: just as laws need precise provisions rather than merely “the general meaning”.

Convergence and Divergence

  • Convergent: the sequence has a limit
  • Divergent: the sequence has no limit

Examples of Convergence

Example 1: lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0

Proof (using the ε-N definition):

For any ε>0\varepsilon > 0, to make ∣1n−0∣<ε|\frac{1}{n} - 0| < \varepsilon, i.e., 1n<ε\frac{1}{n} < \varepsilon, it suffices to take n>1εn > \frac{1}{\varepsilon}.

Take N=⌈1ε⌉N = \lceil \frac{1}{\varepsilon} \rceil (the smallest integer greater than 1ε\frac{1}{\varepsilon}). Then when n>Nn > N, ∣1n−0∣<ε|\frac{1}{n} - 0| < \varepsilon.

Therefore lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0.

Example 2: lim⁡n→∞2n+1n=2\lim_{n \to \infty} \frac{2n+1}{n} = 2

2n+1n=2+1n→2\frac{2n+1}{n} = 2 + \frac{1}{n} \to 2

Examples of Divergence

Example 1: an=na_n = n diverges (tends to infinity)

Example 2: an=(−1)na_n = (-1)^n diverges (oscillates)

Example 3: an=sin⁡na_n = \sin n diverges (irregular oscillation)

Practice Problems

Exercise 1

Use the ε-N definition to prove: lim⁡n→∞1n2=0\lim_{n \to \infty} \frac{1}{n^2} = 0

limit definition of sequences

Approach: Use the ε\varepsilon-NN definition: first solve 1n2<ε\frac{1}{n^2} < \varepsilon for NN, then show by the definition that the requirement holds whenever n>Nn > N.

Steps:

Proof:

For any ε>0\varepsilon > 0, to make ∣1n2−0∣<ε|\frac{1}{n^2} - 0| < \varepsilon, i.e., 1n2<ε\frac{1}{n^2} < \varepsilon, it suffices to have n2>1εn^2 > \frac{1}{\varepsilon}, i.e., n>1εn > \frac{1}{\sqrt{\varepsilon}}.

Take N=⌈1ε⌉N = \lceil \frac{1}{\sqrt{\varepsilon}} \rceil. Then when n>Nn > N, ∣1n2−0∣<ε|\frac{1}{n^2} - 0| < \varepsilon.

Therefore lim⁡n→∞1n2=0\lim_{n \to \infty} \frac{1}{n^2} = 0.

Answer: lim⁡n→∞1n2=0\lim_{n \to \infty} \frac{1}{n^2} = 0.

Exercise 2

Determine whether the following sequences converge. If they converge, find their limits:

  1. an=3n+2n+1a_n = \frac{3n+2}{n+1}
  2. an=n2n+1a_n = \frac{n^2}{n+1}
  3. an=(−1)nna_n = \frac{(-1)^n}{n}
limit definition of sequences

Approach: Judge the limit of each term: when numerator and denominator have the same degree the limit is their ratio, when the degrees differ the higher-degree term dominates, and then draw a conclusion for each item.

Steps：

  1. an=3n+2n+1=3+2n1+1n→31=3a_n = \frac{3n+2}{n+1} = \frac{3 + \frac{2}{n}}{1 + \frac{1}{n}} \to \frac{3}{1} = 3, convergent, with limit 3

  2. an=n2n+1=n1+1n→∞a_n = \frac{n^2}{n+1} = \frac{n}{1 + \frac{1}{n}} \to \infty, divergent

  3. an=(−1)nna_n = \frac{(-1)^n}{n}: since ∣an∣=1n→0|a_n| = \frac{1}{n} \to 0, we have an→0a_n \to 0, convergent, with limit 0

Answer: The three items in turn: an=3n+2n+1a_n = \frac{3n+2}{n+1} converges with limit 33; an=n2n+1a_n = \frac{n^2}{n+1} diverges (it tends to infinity); an=(−1)nna_n = \frac{(-1)^n}{n} converges with limit 00.

Exercise 3

Prove: if lim⁡n→∞an=A\lim_{n \to \infty} a_n = A, then lim⁡n→∞∣an∣=∣A∣\lim_{n \to \infty} |a_n| = |A|

limit definition of sequences

Approach: First rewrite the absolute-value question using ∣an−A∣=∣∣an∣−∣A∣∣|a_n - A| = ||a_n| - |A||, then apply the product and sum rules for limits of sequences.

Steps:

Proof:

From lim⁡n→∞an=A\lim_{n \to \infty} a_n = A, for any ε>0\varepsilon > 0, there exists an NN such that when n>Nn > N, ∣an−A∣<ε|a_n - A| < \varepsilon.

Using the inequality ∣∣an∣−∣A∣∣≤∣an−A∣||a_n| - |A|| \leq |a_n - A| (the triangle inequality for absolute values),

when n>Nn > N, ∣∣an∣−∣A∣∣≤∣an−A∣<ε||a_n| - |A|| \leq |a_n - A| < \varepsilon.

Therefore lim⁡n→∞∣an∣=∣A∣\lim_{n \to \infty} |a_n| = |A|.

Answer: lim⁡n→∞∣an∣=∣A∣\lim_{n \to \infty} |a_n| = |A|.

Summary

Symbols Used in This Article

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡\limOperatorlimitLimit
∞\inftySymbolinfinityInfinity
ε\varepsilonGreek letterepsilonAn arbitrarily small positive number
NNVariableNA certain positive integer
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