Learning Objectives

By the end of this course, you will be able to:

Define a sequence as a function f: ℕ → ℝ and understand the notation
Master the precise ε-N definition of sequence limits
Understand equivalent formulations using neighborhoods and finite terms
Apply the ε-N definition to prove basic limits rigorously
Define and identify divergent sequences using the negation of convergence
Distinguish between bounded divergent and unbounded divergent sequences

Section 1: Definition of a Sequence

Understanding sequences as functions from natural numbers to real numbers

Definition 1.1: Sequence

A sequence is a function $f: \mathbb{N} \to \mathbb{R}$ . We denote the sequence by $\{x_n\}$ where $x_n = f(n)$ , written as:

\{x_n\}: x_1, x_2, \ldots, x_n, \ldots

Key Insights

Domain: $\mathbb{N}$

Natural numbers $\{1, 2, 3, \ldots\}$ representing term indices

Codomain: $\mathbb{R}$

Each term $x_n$ is a real number

Function View

A sequence is fundamentally a function with ordered domain

Historical Context

The rigorous definition of sequences emerged in the 19th century through the work of Augustin-Louis Cauchy (1789-1857) and Karl Weierstrass (1815-1897), who sought to place calculus on firm logical foundations.

Section 2: Examples of Sequences

Sequence	Expression	Behavior	Classification
$\left\{\frac{1}{n}\right\}$	$1, \frac{1}{2}, \frac{1}{3}, \ldots$	Converges to 0	Convergent
$\{(-1)^n\}$	$-1, 1, -1, 1, \ldots$	Oscillates between -1 and 1	Divergent (bounded)
$\{n^2\}$	$1, 4, 9, 16, \ldots$	Tends to +∞	Divergent (unbounded)
$\left\{1 + \frac{(-1)^n}{n}\right\}$	$0, \frac{3}{2}, \frac{2}{3}, \frac{5}{4}, \ldots$	Converges to 1	Convergent

Example 1.1: Newton's Method for √2

Consider the recursive sequence defined by $x_1 = 1$ and $x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$ . This is Newton's method for finding $\sqrt{2}$ .

Step 1: $x_2 = \frac{1}{2}(1 + 2) = 1.5$

Step 2: $x_3 = \frac{1}{2}(1.5 + \frac{2}{1.5}) = 1.41\overline{6}$

Step 3: $x_4 \approx 1.41421568...$

Converges rapidly to $\sqrt{2} \approx 1.41421356...$

Section 3: The ε-N Definition of Limit

The precise mathematical formulation of sequence convergence

Definition 1.2: ε-N Definition of Sequence Limit

Let $\{x_n\}$ be a sequence and $a \in \mathbb{R}$ . We say the sequence $\{x_n\}$ converges to $a$ if:

\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n > N: |x_n - a| < \epsilon

We write $\lim_{n \to \infty} x_n = a$ or $x_n \to a$ as $n \to \infty$ .

Component Analysis

Symbol	Meaning	Interpretation
$\forall \epsilon > 0$	For every positive ε	No matter how small the tolerance
$\exists N \in \mathbb{N}$	There exists a natural number N	We can find a threshold
$\forall n > N$	For all n greater than N	From term N+1 onward
$\|x_n - a\| < \epsilon$	Distance less than ε	Terms stay within ε of a

Remark 1.1: Geometric Interpretation

For any horizontal band of width $2\epsilon$ centered at $y = a$ , all but finitely many terms of the sequence lie within this band. Imagine drawing a strip of height $2\epsilon$ around the limit line—eventually, all sequence points must stay inside this strip, no matter how narrow we make it.

Section 4: Neighborhood Formulation

Definition 1.3: Neighborhood

The ε-neighborhood of $a$ is the open interval:

U(a, \epsilon) = (a - \epsilon, a + \epsilon) = \{x \in \mathbb{R} : |x - a| < \epsilon\}

Equivalent Definition using Neighborhoods

\lim_{n \to \infty} x_n = a \Leftrightarrow \forall \epsilon > 0, \exists N, \forall n > N: x_n \in U(a, \epsilon)

This formulation emphasizes that "being close to $a$ " means "lying in every neighborhood of $a$ ".

Definition 1.4: Finite Terms Outside

$\lim_{n \to \infty} x_n = a$ if and only if for every $\epsilon > 0$ , only finitely many terms of $\{x_n\}$ lie outside $U(a, \epsilon)$ .

Proof of Equivalence of Definitions 1.2 and 1.4:

(1.2) ⟹ (1.4): If all terms from $N+1$ onward lie in $U(a, \epsilon)$ , then at most the first $N$ terms lie outside, which is finite.

(1.4) ⟹ (1.2): If only finitely many terms lie outside $U(a, \epsilon)$ , say $\{x_{n_1}, x_{n_2}, \ldots, x_{n_K}\}$ , take $N = \max\{n_1, n_2, \ldots, n_K\}$ . Then for $n > N$ , we have $x_n \in U(a, \epsilon)$ .

∎

Section 5: Definition of Divergence

Definition 1.5: Not Converging to a Specific Value

$\lim_{n \to \infty} x_n \neq a$ (the sequence does NOT converge to $a$ ) if and only if:

\exists \epsilon_0 > 0, \forall N > 0, \exists n_0 \in \mathbb{N}, n_0 \geq N: |x_{n_0} - a| \geq \epsilon_0

This is the logical negation of the ε-N definition. It says: there exists a "bad" ε₀ such that infinitely many terms escape the ε₀-neighborhood.

Definition 1.6: Divergent Sequence

A sequence $\{x_n\}$ is divergent if it does not converge to any real number:

\forall a \in \mathbb{R}, \lim_{n \to \infty} x_n \neq a

Examples of Divergent Sequences

\{(-1)^n\}

Oscillates between -1 and 1

Bounded divergent

\{\sin n\}

Bounded but irregular oscillation

Bounded divergent

\{n\}

Unbounded, tends to +∞

Unbounded divergent

\{(-1)^n \cdot n\}

Unbounded, oscillates

Unbounded divergent

Key Insight

"Divergent" and "unbounded" are independent concepts. A divergent sequence can be bounded (like $\{(-1)^n\}$ ), and an unbounded sequence is necessarily divergent.

Section 6: Proving Limits Using the Definition

Theorem 1.1: Power Decay

$\lim_{n \to \infty} \frac{1}{n^\alpha} = 0$ for $\alpha > 0$ .

Proof of Theorem 1.1:

Given $\epsilon > 0$ , we need $\frac{1}{n^\alpha} < \epsilon$ , which means $n > \left(\frac{1}{\epsilon}\right)^{1/\alpha}$ .

Take $N = \left\lfloor \left(\frac{1}{\epsilon}\right)^{1/\alpha} \right\rfloor + 1$ .

Then for $n > N$ : $\frac{1}{n^\alpha} < \frac{1}{N^\alpha} < \epsilon$ .

∎

Theorem 1.2: Geometric Decay

$\lim_{n \to \infty} q^n = 0$ for $|q| < 1$ .

Proof of Theorem 1.2:

Given $\epsilon > 0$ , we need $|q|^n < \epsilon$ .

Taking logarithms: $n \ln|q| < \ln \epsilon$ . Since $\ln|q| < 0$ , we get $n > \frac{\ln \epsilon}{\ln |q|}$ .

Take $N = \left\lfloor \frac{\ln \epsilon}{\ln |q|} \right\rfloor + 1$ .

∎

Example 1.2: Rational Function Limit

Prove: $\lim_{n \to \infty} \frac{n^2 - n + 2}{3n^2 + 2n + 4} = \frac{1}{3}$

Solution:

First, compute the difference:

\left|\frac{n^2 - n + 2}{3n^2 + 2n + 4} - \frac{1}{3}\right| = \left|\frac{3(n^2 - n + 2) - (3n^2 + 2n + 4)}{3(3n^2 + 2n + 4)}\right| = \left|\frac{-5n + 2}{9n^2 + 6n + 12}\right|

For large $n$ :

\frac{5n - 2}{9n^2 + 6n + 12} < \frac{5n}{9n^2} = \frac{5}{9n}

Given $\epsilon > 0$ , take $N = \left\lfloor \frac{5}{9\epsilon} \right\rfloor + 1$ . Then for $n > N$ , the expression is less than $\epsilon$ .

Example 1.3: Proving Non-Convergence

Prove: $\lim_{n \to \infty} \left(1 + (-1)^n - \frac{5}{n}\right) \neq 0$

Proof:

Choose $\epsilon_0 = \frac{1}{2}$ . Consider the subsequence $n_k = 2k + 6$ (even terms).

x_{n_k} = 1 + 1 - \frac{5}{2k+6} = 2 - \frac{5}{2k+6}

For $k \geq 1$ : $|x_{n_k} - 0| = 2 - \frac{5}{2k+6} > \frac{7}{6} > 1 > \epsilon_0$

For any $N$ , we can choose $k$ such that $n_k = 2k + 6 > N$ . Then $|x_{n_k}| > \epsilon_0$ .

Theorem 1.3: Constant Sequence

If $x_n = c$ for all $n$ , then $\lim_{n \to \infty} x_n = c$ .

Proof of Theorem 1.3:

Given any $\epsilon > 0$ , for all $n \in \mathbb{N}$ :

|x_n - c| = |c - c| = 0 < \epsilon

We can take $N = 1$ . The condition is satisfied trivially for all $n > N$ .

∎

Theorem 1.4: Shifted Sequence

If $\lim_{n \to \infty} x_n = a$ , then $\lim_{n \to \infty} x_{n+k} = a$ for any fixed $k \in \mathbb{N}$ .

Proof of Theorem 1.4:

Given $\epsilon > 0$ , since $x_n \to a$ , $\exists N$ such that $n > N \Rightarrow |x_n - a| < \epsilon$ .

For the shifted sequence, if $m > N$ , then $m + k > N$ , so $|x_{m+k} - a| < \epsilon$ .

Thus the same $N$ works for the shifted sequence.

∎

Example 1.4: Square Root Sequence

Prove: $\lim_{n \to \infty} \frac{\sqrt{n+1} - \sqrt{n}}{1} = 0$

Solution:

Rationalize the numerator:

\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}

Since $\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}$ , we have:

0 < \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}

Given $\epsilon > 0$ , take $N = \left\lfloor \frac{1}{4\epsilon^2} \right\rfloor + 1$ .

For $n > N$ : $\frac{1}{2\sqrt{n}} < \frac{1}{2\sqrt{N}} < \epsilon$ .

Example 1.5: Factorial Growth

Prove: $\lim_{n \to \infty} \frac{n!}{n^n} = 0$

Solution:

Note that:

\frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}

Each factor $\frac{k}{n} \leq 1$ , and $\frac{1}{n} \leq \frac{1}{n}$ , so:

0 < \frac{n!}{n^n} \leq \frac{1}{n} \to 0

By the Squeeze theorem (covered in CALC-2.3), the limit is 0.

Remark 1.2: Common Proof Strategy

When proving $\lim_{n \to \infty} x_n = a$ using the ε-N definition:

Start with the inequality $|x_n - a| < \epsilon$
Simplify $|x_n - a|$ and find an upper bound in terms of $n$
Solve for $n$ to make the bound less than $\epsilon$
Choose $N$ based on the solution (often $N = \lfloor \text{expression} \rfloor + 1$ )

Key Techniques Summary

For Rational Functions

Divide numerator and denominator by the highest power of $n$ , then use $\frac{1}{n^k} \to 0$ .

For Products

If one factor is bounded and the other tends to zero, the product tends to zero.

For Radicals

Rationalize by multiplying by the conjugate. This often simplifies the expression.

For Divergence

Find a specific $\epsilon_0$ and show infinitely many terms escape the $\epsilon_0$ -neighborhood.

Section 7: Bounded Sequences

Definition 1.5: Bounded Sequence

A sequence $\{x_n\}$ is:

Bounded above if $\exists M \in \mathbb{R}$ : $x_n \leq M$ for all $n$
Bounded below if $\exists m \in \mathbb{R}$ : $x_n \geq m$ for all $n$
Bounded if it is both bounded above and below, i.e., $\exists K > 0$ : $|x_n| \leq K$ for all $n$

Theorem 1.5: Convergent Sequences are Bounded

Every convergent sequence is bounded.

Proof of Theorem 1.5:

Let $\lim_{n \to \infty} x_n = L$ . Take $\epsilon = 1$ . Then $\exists N$ such that for $n > N$ :

|x_n - L| < 1 \Rightarrow |x_n| < |L| + 1

The first $N$ terms $x_1, x_2, \ldots, x_N$ form a finite set.

Let $K = \max\{|x_1|, |x_2|, \ldots, |x_N|, |L| + 1\}$ .

Then $|x_n| \leq K$ for all $n \in \mathbb{N}$ .

∎

Remark 1.3: Converse is False

The converse is FALSE: bounded does NOT imply convergent. The sequence $\{(-1)^n\}$ is bounded ( $|(-1)^n| = 1 \leq 1$ for all $n$ ) but diverges because it oscillates between -1 and 1.

Example 1.6: Bounded but Divergent

Show: $\{\sin n\}$ is bounded but divergent.

Bounded: $|\sin n| \leq 1$ for all $n$ .

Divergent: The sequence does not converge because $\sin n$ takes values dense in $[-1, 1]$ (since $\pi$ is irrational, $n \mod 2\pi$ is dense in $[0, 2\pi)$ ).

Relationship Summary

Convergent ⟹ Bounded (always true)

Bounded ⟹ Convergent (FALSE in general)

Bounded + Monotone ⟹ Convergent (Monotone Bounded Theorem, CALC-2.3)

Section 7: Practice Problems

Practice proving limits using the ε-N definition

Recommended Practice

1. Basic Limits

Prove $\lim_{n \to \infty} \frac{1}{n} = 0$ directly from the ε-N definition.

2. Rational Functions

Prove $\lim_{n \to \infty} \frac{3n+1}{2n+5} = \frac{3}{2}$ .

3. Finding N explicitly

For $x_n = \frac{n}{n+1}$ and $\epsilon = 0.001$ , find the smallest valid N.

Section 8: Advanced Examples

Example 1.7: Limit of Geometric Sequence

Problem: Prove $\lim_{n \to \infty} r^n = 0$ for $|r| < 1$ .

Proof:

Given $\epsilon > 0$ . Since $|r| < 1$ , we can write $|r| = \frac{1}{1+h}$ where $h > 0$ .

By Bernoulli's inequality: $(1+h)^n \geq 1 + nh$ .

|r|^n = \frac{1}{(1+h)^n} \leq \frac{1}{1 + nh}

Choose $N > \frac{1 - \epsilon}{h\epsilon}$ . For $n > N$ :

|r^n - 0| = |r|^n \leq \frac{1}{1 + nh} < \frac{1}{nh} < \epsilon

Example 1.8: Root Sequence

Problem: Prove $\lim_{n \to \infty} \sqrt[n]{a} = 1$ for $a > 1$ .

Proof:

Let $h_n = \sqrt[n]{a} - 1$ . Since $a > 1$ , we have $h_n > 0$ .

Then $a = (1 + h_n)^n$ . By binomial theorem:

a = (1 + h_n)^n \geq 1 + nh_n > nh_n

So $h_n < \frac{a}{n}$ .

Given $\epsilon > 0$ , choose $N > \frac{a}{\epsilon}$ . For $n > N$ :

|\sqrt[n]{a} - 1| = h_n < \frac{a}{n} < \epsilon

Example 1.9: Squeeze Applied

Problem: Find $\lim_{n \to \infty} \frac{\cos n}{n^2}$ .

Solution:

Since $-1 \leq \cos n \leq 1$ , we have:

-\frac{1}{n^2} \leq \frac{\cos n}{n^2} \leq \frac{1}{n^2}

Both $\frac{-1}{n^2} \to 0$ and $\frac{1}{n^2} \to 0$ .

By Squeeze Theorem: $\lim_{n \to \infty} \frac{\cos n}{n^2} = 0$ .

Theorem 1.6: Algebra of Limits

If $\lim x_n = a$ and $\lim y_n = b$ , then:

$\lim(x_n + y_n) = a + b$
$\lim(x_n \cdot y_n) = a \cdot b$
$\lim\frac{x_n}{y_n} = \frac{a}{b}$ (if $b \neq 0$ )
$\lim(cx_n) = ca$ for any constant $c$

Proof of Theorem 1.6 (Sum Rule):

Given $\epsilon > 0$ . Since $x_n \to a$ , $\exists N_1$ such that $|x_n - a| < \frac{\epsilon}{2}$ for $n > N_1$ .

Since $y_n \to b$ , $\exists N_2$ such that $|y_n - b| < \frac{\epsilon}{2}$ for $n > N_2$ .

Let $N = \max(N_1, N_2)$ . For $n > N$ :

|(x_n + y_n) - (a + b)| \leq |x_n - a| + |y_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

∎

Remark 1.5: Common Mistakes

Mistake 1: Using N that depends on n. The value N must be a fixed number.

Mistake 2: Claiming $|x_n - a| \leq \epsilon$ without specifying for which n.

Correct: Always state "for all $n > N$ " after specifying N.

Key Takeaways

The ε-N definition is the rigorous foundation of limits
Limits, if they exist, are unique
Convergent sequences are bounded
The negation helps prove divergence

Example 1.10: Limit with Square Root

Problem: Prove $\lim_{n \to \infty} \frac{\sqrt{n+1} - \sqrt{n}}{1} = 0$ .

Solution:

Rationalize: $\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$

Since $\sqrt{n+1} + \sqrt{n} > \sqrt{n}$ :

0 < \sqrt{n+1} - \sqrt{n} < \frac{1}{\sqrt{n}} \to 0

By Squeeze Theorem, $\sqrt{n+1} - \sqrt{n} \to 0$ .

Theorem 1.7: Monotone Subsequence Lemma

Every sequence has a monotone subsequence (increasing or decreasing).

Remark 1.6: Proof Strategy Summary

To prove convergence:

Find candidate limit $a$
Express $|x_n - a|$ simply
Make it $< \epsilon$ for $n > N$

To prove divergence:

Find $\epsilon_0 > 0$
For any $N$ , find $n_0 > N$
Show $|x_{n_0} - a| \geq \epsilon_0$

Example 1.11: Sequence with Alternating Signs

Problem: Prove $\lim_{n \to \infty} \frac{(-1)^n}{n} = 0$ .

Solution:

\left|\frac{(-1)^n}{n} - 0\right| = \frac{|(-1)^n|}{n} = \frac{1}{n}

Given $\epsilon > 0$ , choose $N > \frac{1}{\epsilon}$ .

For $n > N$ : $\frac{1}{n} < \frac{1}{N} < \epsilon$ . ∎

Theorem 1.8: Arithmetic Mean Convergence

If $\lim_{n \to \infty} x_n = L$ , then:

\lim_{n \to \infty} \frac{x_1 + x_2 + \cdots + x_n}{n} = L

Note: The converse is false. Example: $x_n = (-1)^n$ has arithmetic mean converging to 0.

Example 1.12: Cesàro Mean Application

Problem: If $x_n = 1 + \frac{1}{n}$ , find $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} x_k$ .

Solution:

Since $x_n \to 1$ , by Cesàro mean theorem:

\frac{1}{n}\sum_{k=1}^{n} x_k \to 1

Direct calculation: $\frac{1}{n}\sum_{k=1}^{n}\left(1 + \frac{1}{k}\right) = 1 + \frac{H_n}{n}$ where $H_n \sim \ln n$ .

Remark 1.7: Important Limit Patterns

Polynomial/Polynomial

$\frac{a_m n^m + \cdots}{b_k n^k + \cdots} \to \begin{cases} 0 & m < k \\ \frac{a_m}{b_k} & m = k \\ \pm\infty & m > k \end{cases}$

Exponential Form

$\left(1 + \frac{a}{n}\right)^{bn} \to e^{ab}$

Common Pitfalls to Avoid

✗Assuming $x_n < a_n$ implies $\lim x_n < \lim a_n$ (only ≤ holds)
✗Forgetting that N depends on ε in the definition
✗Using limit laws when limits don't exist

Practice Quiz: Sequences and ε-N Definition

8

Questions

0

Correct

0%

Accuracy

1

Which of the following sequences is convergent?

Easy

Not attempted

2

The ε-N definition states:

\lim_{n \to \infty} x_n = a

if and only if...

Easy

Not attempted

3

For the sequence

x_n = \frac{1}{n}

with limit 0 and

\epsilon = 0.01

, what is the smallest valid N?

Medium

Not attempted

4

To prove

\lim_{n \to \infty} \frac{2n+1}{n+3} = 2

, we compute

|\frac{2n+1}{n+3} - 2|

. This equals:

Medium

Not attempted

5

A sequence

\{x_n\}

does NOT converge to

a

if and only if:

Hard

Not attempted

6

Prove that

\{(-1)^n\}

does not converge to 0. Which ε₀ works?

Medium

Not attempted

7

Consider the recursive sequence

x_1 = 1

,

x_{n+1} = \frac{1}{2}(x_n + \frac{2}{x_n})

. If it converges to L, what is L?

Hard

Not attempted

8

If only finitely many terms of a sequence lie outside every neighborhood of

a

, then:

Hard

Not attempted

Frequently Asked Questions

Why do we need such a precise definition of limits?

Before the ε-N definition, mathematicians used vague terms like 'approaches' or 'gets arbitrarily close.' This led to paradoxes and incorrect proofs. The ε-N definition provides a rigorous framework that eliminates ambiguity and allows us to prove statements about limits with certainty.

What's the intuition behind the ε-N definition?

Think of ε as a 'tolerance level' - how close we want the sequence to be to the limit. N is the 'starting point' after which all terms are within this tolerance. The definition says: no matter how tight we make the tolerance (ε), we can always find a point (N) after which all terms satisfy it.

Does the choice of N depend on ε?

Yes! For smaller values of ε, we typically need a larger N. For example, for the sequence 1/n → 0, if ε = 0.1, we need N = 10. If ε = 0.01, we need N = 100. The key is that such an N must exist for ANY positive ε.

Can a sequence have more than one limit?

No! One of the fundamental theorems (which we prove in CALC-2.2) shows that if a sequence converges, its limit is unique. This is why we can write 'the limit' instead of 'a limit.'

What's the difference between 'bounded' and 'convergent'?

Convergent implies bounded, but the converse is false. The sequence {(-1)^n} is bounded (all terms are between -1 and 1) but divergent because it oscillates forever. However, every convergent sequence must be bounded.

How do I choose the right N in a proof?

Work backwards: start with |xₙ - a| < ε and algebraically solve for what condition on n makes this true. The resulting expression tells you how to choose N. Common techniques include using floor functions or adding 1 to ensure N is a natural number.

Sequences and the ε-N Definition

Section 1: Definition of a Sequence

Domain: N\mathbb{N}N

Codomain: R\mathbb{R}R

Function View

Historical Context

Section 2: Examples of Sequences

Section 3: The ε-N Definition of Limit

Section 4: Neighborhood Formulation

Section 5: Definition of Divergence

Key Insight

Section 6: Proving Limits Using the Definition

For Rational Functions

For Products

For Radicals

For Divergence

Section 7: Bounded Sequences

Section 7: Practice Problems

1. Basic Limits

2. Rational Functions

3. Finding N explicitly

Section 8: Advanced Examples

To prove convergence:

To prove divergence:

Frequently Asked Questions

Domain: $\mathbb{N}$

Codomain: $\mathbb{R}$