Convergence of Sequences

The concept of convergence is central to analysis. A sequence converges if its terms approach a single limiting value as the index tends to infinity. The epsilon-delta definition, formalized by Cauchy and Weierstrass in the 19th century, provides a rigorous foundation for calculus and analysis.

Definition and basic properties

Definition2.1Convergence of a sequence

A sequence $(a_n)$ in $\mathbb{R}$ converges to a limit $L \in \mathbb{R}$ if for every $\epsilon > 0$ , there exists $N \in \mathbb{N}$ such that

$|a_n - L| < \epsilon \quad \text{for all } n \geq N.$

We write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$ .

RemarkEpsilon-N definition

The definition says: eventually, all terms of the sequence lie within any prescribed distance $\epsilon$ of the limit $L$ . No matter how small $\epsilon > 0$ is chosen, only finitely many terms lie outside the interval $(L - \epsilon, L + \epsilon)$ .

ExampleConvergence of 1/n

Let $a_n = 1/n$ . Then $a_n \to 0$ as $n \to \infty$ .

Proof: Given $\epsilon > 0$ , by the Archimedean property, choose $N \in \mathbb{N}$ such that $1/N < \epsilon$ . Then for all $n \geq N$ ,

$|a_n - 0| = \frac{1}{n} \leq \frac{1}{N} < \epsilon.$

Thus $a_n \to 0$ .

ExampleConstant sequences

If $a_n = c$ for all $n$ (constant), then $a_n \to c$ . Indeed, for any $\epsilon > 0$ , $|a_n - c| = 0 < \epsilon$ for all $n \geq 1$ .

ExampleAlternating sequence

Let $a_n = (-1)^n/n$ . Then $a_n \to 0$ . For $|a_n - 0| = 1/n \to 0$ as $n \to \infty$ , regardless of the sign.

ExampleDivergent sequence

Let $a_n = n$ . Then $(a_n)$ does not converge. For any proposed limit $L$ , taking $\epsilon = 1$ , we have $|a_n - L| = |n - L| \to \infty$ , so no $N$ works. We write $a_n \to +\infty$ (diverges to infinity).

Uniqueness of limits

Theorem2.1Uniqueness of limits

If $a_n \to L$ and $a_n \to L'$ , then $L = L'$ .

Proof

Suppose $L \neq L'$ . Let $\epsilon = |L - L'|/2 > 0$ . Since $a_n \to L$ , there exists $N_1$ such that $|a_n - L| < \epsilon$ for all $n \geq N_1$ . Similarly, there exists $N_2$ such that $|a_n - L'| < \epsilon$ for all $n \geq N_2$ . Let $N = \max(N_1, N_2)$ . Then for $n \geq N$ ,

$|L - L'| = |(L - a_n) + (a_n - L')| \leq |L - a_n| + |a_n - L'| < \epsilon + \epsilon = |L - L'|,$

a contradiction. Thus $L = L'$ .

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Algebra of limits

Theorem2.2Algebra of limits

If $a_n \to a$ and $b_n \to b$ , then:

$a_n + b_n \to a + b$ .
$a_n - b_n \to a - b$ .
$a_n b_n \to ab$ .
If $b \neq 0$ , then $a_n / b_n \to a/b$ (assuming $b_n \neq 0$ for all $n$ ).

RemarkProof sketch

These are proved using the epsilon-N definition and the triangle inequality. For example, for (1): given $\epsilon > 0$ , choose $N_1$ such that $|a_n - a| < \epsilon/2$ for $n \geq N_1$ , and $N_2$ such that $|b_n - b| < \epsilon/2$ for $n \geq N_2$ . For $n \geq \max(N_1, N_2)$ ,

$|(a_n + b_n) - (a + b)| \leq |a_n - a| + |b_n - b| < \epsilon/2 + \epsilon/2 = \epsilon.$

ExampleSum of convergent sequences

Let $a_n = 1/n$ and $b_n = 1/n^2$ . Then $a_n \to 0$ , $b_n \to 0$ , and $a_n + b_n = 1/n + 1/n^2 \to 0 + 0 = 0$ .

ExampleProduct of sequences

Let $a_n = 1 + 1/n \to 1$ and $b_n = 2 - 1/n \to 2$ . Then $a_n b_n = (1 + 1/n)(2 - 1/n) = 2 + 1/n - 1/n^2 \to 1 \cdot 2 = 2$ .

Squeeze theorem

Theorem2.3Squeeze Theorem

Suppose $a_n \leq b_n \leq c_n$ for all $n \geq N_0$ , and $a_n \to L$ , $c_n \to L$ . Then $b_n \to L$ .

Proof

Given $\epsilon > 0$ , choose $N_1 \geq N_0$ such that $|a_n - L| < \epsilon$ for $n \geq N_1$ , i.e., $L - \epsilon < a_n < L + \epsilon$ . Similarly, choose $N_2$ such that $L - \epsilon < c_n < L + \epsilon$ for $n \geq N_2$ . For $n \geq \max(N_1, N_2)$ ,

$L - \epsilon < a_n \leq b_n \leq c_n < L + \epsilon,$

so $|b_n - L| < \epsilon$ . Thus $b_n \to L$ .

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ExampleSqueeze theorem application

Let $b_n = \sin(n)/n$ . Since $-1 \leq \sin(n) \leq 1$ , we have

$-\frac{1}{n} \leq \frac{\sin(n)}{n} \leq \frac{1}{n}.$

Since $1/n \to 0$ and $-1/n \to 0$ , by the squeeze theorem, $\sin(n)/n \to 0$ .

Bounded sequences

Definition2.2Bounded sequence

A sequence $(a_n)$ is bounded if there exists $M > 0$ such that $|a_n| \leq M$ for all $n \in \mathbb{N}$ .

Theorem2.4Convergent sequences are bounded

If $a_n \to L$ , then $(a_n)$ is bounded.

Proof

Choose $\epsilon = 1$ . There exists $N$ such that $|a_n - L| < 1$ for all $n \geq N$ , i.e., $L - 1 < a_n < L + 1$ for $n \geq N$ . Let

$M = \max(|a_1|, |a_2|, \ldots, |a_{N-1}|, |L| + 1).$

Then $|a_n| \leq M$ for all $n \in \mathbb{N}$ .

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RemarkConverse is false

A bounded sequence need not converge. For example, $a_n = (-1)^n$ is bounded ( $|a_n| = 1$ ) but does not converge (it oscillates between $-1$ and $1$ ).

ExampleBounded divergent sequence

Let $a_n = \sin(n)$ . Then $|a_n| \leq 1$ for all $n$ , so $(a_n)$ is bounded. However, $(a_n)$ does not converge: the sequence $\sin(1), \sin(2), \sin(3), \ldots$ is dense in $[-1, 1]$ and has no limit.

Subsequences and Bolzano-Weierstrass

Definition2.3Subsequence

A subsequence of $(a_n)$ is a sequence $(a_{n_k})$ where $n_1 < n_2 < n_3 < \cdots$ is a strictly increasing sequence of indices.

Theorem2.5Bolzano-Weierstrass Theorem

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

RemarkProof deferred

The proof uses the completeness of $\mathbb{R}$ and the nested intervals property. See Bolzano-Weierstrass Theorem for details.

ExampleExtracting a convergent subsequence

Let $a_n = (-1)^n$ . The sequence does not converge, but the subsequence $a_{2k} = 1$ converges to $1$ , and $a_{2k+1} = -1$ converges to $-1$ . Both are convergent subsequences.

Summary

Convergence is the foundation of real analysis:

A sequence converges if its terms eventually lie within any $\epsilon$ of the limit.
Limits are unique, and arithmetic operations preserve limits.
The squeeze theorem is a powerful tool for computing limits.
Convergent sequences are bounded, but not conversely.
Bolzano-Weierstrass ensures every bounded sequence has a convergent subsequence.

See Cauchy Sequences for an alternative characterization of convergence, and Monotone Convergence Theorem for a major application of completeness.