---
title: Sequences and Their Limits
module: Sequences and Series
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  A sequence is a function on the natural numbers; it converges to a limit when
  its terms eventually stay within any prescribed tolerance of that number. The
  epsilon-M definition fixes the order of the quantifiers, and from it the limit
  is unique, every convergent sequence is bounded, and only the tail matters.
  Divergence to plus or minus infinity records terms that outgrow every bound.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "Ch. 2 — Sequences and Series; §2.1 Sequences and limits"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

Analysis is the study of limits, and the simplest is the limit of a sequence of
real numbers. Continuity, the derivative, the integral, and the interchange of
limit operations all reduce to the single $\epsilon$-$M$ definition below.

## Sequences

> **Definition (Sequence).** A **sequence** of real numbers is a function
> $x \colon \mathbb{N} \to \mathbb{R}$. We write the $n$th value $x(n)$ as
> $x_n$ and denote the whole sequence by $\{x_n\}_{n=1}^\infty$.

The subscript notation resembles set notation, but a sequence and a set are
different objects. A sequence is ordered and may repeat values; the set of
values it takes, its **range**, forgets both. For example
$\{(-1)^n\}_{n=1}^\infty$ is the sequence $-1, 1, -1, 1, \dots$, an infinite
list, whereas its range is the two-element set $\{-1, 1\}$.

A sequence is usually specified by a formula for its $n$th term. Three that
recur throughout the subject:

- $\{1/n\}_{n=1}^\infty = 1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \dots$, decreasing toward $0$.
- $\{n\}_{n=1}^\infty = 1, 2, 3, \dots$, growing without bound.
- The **constant sequence** $\{c\}_{n=1}^\infty = c, c, c, \dots$ for a fixed $c \in \mathbb{R}$.

> **Definition (Bounded sequence).** A sequence $\{x_n\}_{n=1}^\infty$ is
> **bounded** if there exists $B \in \mathbb{R}$ with $|x_n| \le B$ for all
> $n \in \mathbb{N}$. Equivalently, the range $\{x_n : n \in \mathbb{N}\}$ is a
> bounded set. Bounded above and bounded below are defined the same way with
> one-sided inequalities.

The sequence $\{1/n\}$ is bounded ($B = 1$ works), and $\{n\}$ is not. This
lesson uses the absolute value and its triangle inequality
[from the foundations](/real-analysis/foundations/absolute-value-bounds)
constantly; the estimate $|a + b| \le |a| + |b|$ appears in nearly every proof
below.

## Convergence

The idea is that a sequence **converges** to $x$ if its terms are eventually as
close to $x$ as we like. "As close as we like" is made precise by a tolerance
$\epsilon > 0$; "eventually" is made precise by an index $M$ past which the
tolerance holds.

> **Definition (Convergence).** A sequence $\{x_n\}_{n=1}^\infty$ **converges**
> to $x \in \mathbb{R}$ if for every $\epsilon > 0$ there exists $M \in
> \mathbb{N}$ such that
> $$
> |x_n - x| < \epsilon \qquad \text{for all } n \ge M.
> $$
> The number $x$ is a **limit** of the sequence. A sequence that converges is
> **convergent**; otherwise it **diverges**.

The order of the quantifiers is the whole content of the definition. The
tolerance $\epsilon$ is given first, and only then do we produce $M$, so $M$ is
allowed to depend on $\epsilon$ — a smaller tolerance generally forces a larger
$M$. The condition $|x_n - x| < \epsilon$ says $x_n$ lies in the open interval
$(x - \epsilon,\, x + \epsilon)$. So convergence means: every such interval
around $x$, no matter how narrow, contains all but finitely many terms of the
sequence.

$$
% caption: Convergence to L. Given the band of half-width epsilon around L,
% all terms from index M onward lie inside it; earlier terms may be anywhere.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[black] (0,0) -- (10.2,0);
\draw[black] (0,0) -- (0,4.8);
% band
\fill[acc!10] (0,1.8) rectangle (10.2,4.2);
\draw[acc, thick] (0,3.0) -- (10.2,3.0) node[right, black] {$L$};
\draw[acc!70, dashed] (0,4.2) -- (10.2,4.2);
\draw[acc!70, dashed] (0,1.8) -- (10.2,1.8);
% M marker
\draw[black, dashed] (2.55,0) -- (2.55,4.8);
\node[black, anchor=south] at (2.55,4.8) {index $M$};
% points x_n at y=(v-1)*3, x=n*0.95
\foreach \n/\y in {1/0.0, 2/4.5, 3/2.0, 4/3.75, 5/2.4, 6/3.5, 7/2.57, 8/3.375, 9/2.667, 10/3.3} {
  \fill[black] (\n*0.95,\y) circle (2.4pt);
  \draw[black, thin] (\n*0.95,0) -- (\n*0.95,\y);
}
\node[anchor=north] at (2*0.95,-0.05) {$x_2$};
\node[anchor=north, black] at (9.7,-0.05) {$n$};
\end{tikzpicture}
$$

$$
% caption: The same convergence read on the number line. All but finitely many
% terms fall in the band of half-width epsilon about L; finitely many sit outside.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (11,0);
\fill[acc!12] (4.5,-0.4) rectangle (7.5,0.4);
\draw[acc!70, dashed] (4.5,-0.55) -- (4.5,0.55);
\draw[acc!70, dashed] (7.5,-0.55) -- (7.5,0.55);
\draw[acc, thick] (6,-0.6) -- (6,0.6);
\node[acc, anchor=south] at (6,0.6) {$L$};
\foreach \x in {5.0,5.35,5.7,6.05,6.35,6.65,6.95,5.15}
  \fill[black] (\x,0) circle (2.0pt);
\fill[black] (1.2,0) circle (2.0pt);
\fill[black] (9.6,0) circle (2.0pt);
\fill[black] (2.7,0) circle (2.0pt);
\node[black, anchor=north] at (1.2,-0.12) {$x_1$};
\node[black, anchor=north] at (9.6,-0.12) {$x_2$};
\node[black, anchor=north] at (2.7,-0.12) {$x_3$};
\node[acc, anchor=north] at (6,-0.55) {tail in band};
\end{tikzpicture}
$$

> **Worked example.** The constant sequence $\{c\}_{n=1}^\infty$ converges to
> $c$. For any $\epsilon > 0$, take $M = 1$; then $|x_n - c| = |c - c| = 0 <
> \epsilon$ for all $n$.

> **Worked example.** The sequence $\{1/n\}$ converges to $0$. Given $\epsilon >
> 0$, the [Archimedean property](/real-analysis/foundations/ordered-fields-completeness)
> gives $M \in \mathbb{N}$ with $1/M < \epsilon$; the least such index is
> $M = \lfloor 1/\epsilon \rfloor + 1$. Then for all $n \ge M$,
> $$
> \left| \tfrac{1}{n} - 0 \right| = \tfrac{1}{n} \le \tfrac{1}{M} < \epsilon.
> $$

$$
% caption: The sequence 1/n descending to 0. Past the index M with 1/M < epsilon,
% every term lies in the band of half-width epsilon about 0; earlier terms are above it.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10.4,0) node[right, black] {$n$};
\draw[black] (0,-0.15) -- (0,3.5);
\draw[acc, thick] (0,0.3) -- (10.4,0.3) node[right, black] {$0$};
\fill[acc!10] (0,0.3) rectangle (10.4,0.85);
\draw[acc!70, dashed] (0,0.85) -- (10.4,0.85);
\draw[black, dashed] (5.5,0) -- (5.5,3.4);
\node[black, anchor=south] at (5.5,3.4) {index $M$};
\foreach \n in {1,2,3,4,5,6,7,8,9,10}
  \fill[black] (\n*0.95,{0.3+2.9/\n}) circle (2.0pt);
\node[black, anchor=south west] at (1.0,3.15) {$1$};
\end{tikzpicture}
$$

> **Worked example.** The sequence $\left\{\frac{n^2+1}{n^2+n}\right\}$ converges
> to $1$. Given $\epsilon > 0$, pick $M \in \mathbb{N}$ with $1/M < \epsilon$.
> For all $n \ge M$,
> $$
> \left| \frac{n^2+1}{n^2+n} - 1 \right| = \frac{|1 - n|}{n^2+n} = \frac{n-1}{n^2+n} \le \frac{n}{n^2+n} = \frac{1}{n+1} \le \frac{1}{n} \le \frac{1}{M} < \epsilon.
> $$
> Trading the sharp bound $\frac{n-1}{n^2+n}$ for the looser $\frac{1}{n}$ is
> deliberate: any cruder estimate that still beats $\epsilon$ suffices.

> **Worked example.** The sequence $\{(-1)^n\}$ diverges. Suppose it converged to
> some $x$. Take $\epsilon = \tfrac{1}{2}$ and let $M$ be as in the definition.
> For an even $n \ge M$ we would have both $|1 - x| < \tfrac{1}{2}$ and, at
> $n+1$, $|-1 - x| < \tfrac{1}{2}$. The triangle inequality forces the
> contradiction
> $$
> 2 = |1 - x - (-1 - x)| \le |1 - x| + |-1 - x| < \tfrac{1}{2} + \tfrac{1}{2} = 1.
> $$

The last is the standard way to prove divergence: assume a limit,
choose one specific $\epsilon$ small enough to separate two persistent clusters
of terms, and derive a contradiction from the triangle inequality.

## The limit is unique

We wrote "a limit" above because uniqueness must be proved. Once it is, we may
write $\lim_{n\to\infty} x_n = x$ and speak of **the** limit.

> **Proposition (Uniqueness of limits).** A convergent sequence has exactly one
> limit.

The proof is the prototype of a technique used everywhere in analysis: to show
a fixed nonnegative quantity is zero, bound it above by an arbitrary
$\epsilon > 0$.

> **Proof.** Suppose $\{x_n\}$ has limits $x$ and $y$. Fix $\epsilon > 0$. Choose
> $M_1$ so that $|x_n - x| < \epsilon/2$ for $n \ge M_1$, and $M_2$ so that
> $|x_n - y| < \epsilon/2$ for $n \ge M_2$. For any $n \ge \max\{M_1, M_2\}$,
> $$
> |y - x| = |x_n - x - (x_n - y)| \le |x_n - x| + |x_n - y| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon.
> $$
> Since $|y - x| < \epsilon$ for every $\epsilon > 0$, we must have $|y - x| = 0$,
> so $y = x$.[^lebl-uniq] $\square$

The maneuver of splitting $y - x$ as $(x_n - x) - (x_n - y)$ (inserting and
subtracting a term the sequence gets close to) is called adding and
subtracting inside an absolute value, and it appears in nearly every proof in
this module.

## Convergent sequences are bounded

> **Proposition.** Every convergent sequence is bounded.

> **Proof.** Suppose $x_n \to x$. Apply the definition with $\epsilon = 1$: there
> is $M$ with $|x_n - x| < 1$ for all $n \ge M$. For those $n$,
> $$
> |x_n| = |x_n - x + x| \le |x_n - x| + |x| < 1 + |x|.
> $$
> The finitely many earlier terms are bounded by their maximum, so setting
> $$
> B = \max\bigl\{ |x_1|, |x_2|, \dots, |x_{M-1}|,\ 1 + |x| \bigr\}
> $$
> gives $|x_n| \le B$ for all $n$.[^lebl-bdd] $\square$

The converse fails: $\{(-1)^n\}$ is bounded but divergent. Boundedness is
necessary for convergence, not sufficient; the missing hypothesis is supplied by
monotonicity or by the extraction of a convergent subsequence.

$$
% caption: Three archetypes. A convergent sequence settles into any band around
% its limit; a bounded divergent one oscillates without settling; an unbounded
% one escapes every band.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% panel 1: convergent
\begin{scope}
  \draw[black] (0,0) -- (3.2,0);
  \draw[black] (0,-1.4) -- (0,1.9);
  \draw[acc!70, dashed] (0,0.8) -- (3.2,0.8);
  \draw[acc!70, dashed] (0,1.2) -- (3.2,1.2);
  \foreach \n/\y in {1/-1.2, 2/1.6, 3/0.6, 4/1.35, 5/0.85, 6/1.15, 7/0.95, 8/1.08, 9/1.0, 10/1.04}
    \fill[black] (\n*0.3,\y) circle (1.7pt);
  \node[anchor=north, black] at (1.6,-1.55) {convergent};
\end{scope}
% panel 2: bounded divergent
\begin{scope}[xshift=4.2cm]
  \draw[black] (0,0) -- (3.2,0);
  \draw[black] (0,-1.4) -- (0,1.9);
  \foreach \n in {1,3,5,7,9} \fill[black] (\n*0.3,-1.0) circle (1.7pt);
  \foreach \n in {2,4,6,8,10} \fill[black] (\n*0.3,1.4) circle (1.7pt);
  \node[anchor=north, black] at (1.6,-1.55) {bounded, divergent};
\end{scope}
% panel 3: unbounded
\begin{scope}[xshift=8.4cm]
  \draw[black] (0,0) -- (3.2,0);
  \draw[black] (0,-1.4) -- (0,1.9);
  \foreach \n/\y in {1/0.1, 2/0.35, 3/0.6, 4/0.85, 5/1.1, 6/1.35, 7/1.6, 8/1.75}
    \fill[black] (\n*0.35,\y) circle (1.7pt);
  \node[anchor=north, black] at (1.4,-1.55) {unbounded};
\end{scope}
\end{tikzpicture}
$$

## Convergence depends only on the tail

Changing, deleting, or inserting finitely many terms cannot affect whether a
sequence converges or what it converges to.

> **Definition (Tail).** For a sequence $\{x_n\}_{n=1}^\infty$ and $K \in
> \mathbb{N}$, the **$K$-tail** is the sequence
> $\{x_{n+K}\}_{n=1}^\infty = x_{K+1}, x_{K+2}, x_{K+3}, \dots$, i.e. the
> original sequence with its first $K$ terms discarded.

> **Proposition (Tails).** For a sequence $\{x_n\}$ the following are
> equivalent: (i) $\{x_n\}$ converges; (ii) the $K$-tail converges for every
> $K$; (iii) the $K$-tail converges for some $K$. When any limit exists,
> $\lim_{n\to\infty} x_n = \lim_{n\to\infty} x_{n+K}$ for every $K$.

> **Proof.** _(i) $\Rightarrow$ (ii)._ Suppose $x_n \to x$ and fix $K$. Given
> $\epsilon > 0$, take $M$ with $|x_n - x| < \epsilon$ for $n \ge M$. Since
> $n \ge M$ implies $n + K \ge M$, we get $|x_{n+K} - x| < \epsilon$ for all
> $n \ge M$, so the $K$-tail converges to $x$.
>
> _(iii) $\Rightarrow$ (i)._ Suppose the $K$-tail converges to $x$: given
> $\epsilon > 0$, there is $M'$ with $|x_{n+K} - x| < \epsilon$ for $n \ge M'$.
> Put $M = M' + K$. For $n \ge M$ we have $n - K \ge M'$, hence $|x_n - x| =
> |x_{(n-K)+K} - x| < \epsilon$. Since (ii) $\Rightarrow$ (iii) is trivial, all
> three are equivalent.[^lebl-tail] $\square$

The limit does not depend on how the sequence begins. This licenses arguments
that establish a property "for large $n$" — for instance, a sequence that is
monotone only from some index onward converges as soon as the tail is monotone
and bounded, because the tail governs the limit.

## Divergence to infinity

Some divergent sequences have terms that increase past every bound. It is
convenient to record this behavior with its own notation, even though such
sequences do not converge to a real number.

> **Definition (Divergence to infinity).** A sequence $\{x_n\}$ **diverges to
> infinity**, written $\lim_{n\to\infty} x_n = \infty$, if for every $K \in
> \mathbb{R}$ there is $M \in \mathbb{N}$ such that $x_n > K$ for all $n \ge M$.
> It **diverges to minus infinity**, written $\lim_{n\to\infty} x_n = -\infty$,
> if for every $K \in \mathbb{R}$ there is $M$ with $x_n < K$ for all $n \ge M$.

The structure mirrors ordinary convergence: an arbitrary threshold $K$ is given
first, then an index $M$ past which every term clears it. With this convention,

$$
\lim_{n\to\infty} n = \infty, \qquad \lim_{n\to\infty} n^2 = \infty, \qquad \lim_{n\to\infty} (-n) = -\infty.
$$

A sequence diverging to $\pm\infty$ is unbounded, but the converse fails:
$x_n = 0$ for odd $n$ and $x_n = n$ for even $n$ is unbounded yet does not
diverge to $\infty$, because it does not eventually stay above every threshold.
The finer bookkeeping for such sequences uses the
[limit superior and limit inferior](/real-analysis/sequences-series/limsup-bolzano-weierstrass).

$$
% caption: Divergence to infinity. Every threshold K is eventually cleared: past
% some index M, all terms sit above the line, and this holds for arbitrarily high K.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.5,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,4.4);
% threshold line
\draw[acc, thick] (0,2.6) -- (9.5,2.6) node[right, black] {$K$};
\draw[black, dashed] (4.9,0) -- (4.9,4.4);
\node[black, anchor=south] at (4.9,4.4) {index $M$};
\foreach \n/\y in {1/0.5, 2/0.75, 3/1.05, 4/1.4, 5/1.8, 6/2.25, 7/2.75, 8/3.3, 9/3.9}
  \fill[black] (\n*1.0,\y) circle (2.2pt);
\end{tikzpicture}
$$

## Summary

Convergence is an $\epsilon$-then-$M$ statement; a limit, once shown unique, is
written $\lim_{n\to\infty} x_n$; convergence forces boundedness but not
conversely; and only the tail matters.

| Statement | Meaning | Quantifier order |
| --- | --- | --- |
| $x_n \to x$ | terms eventually within $\epsilon$ of $x$ | $\forall \epsilon\ \exists M\ \forall n \ge M$ |
| $x_n \to \infty$ | terms eventually above any threshold | $\forall K\ \exists M\ \forall n \ge M$ |
| bounded | one bound $B$ holds for all terms | $\exists B\ \forall n$ |
| convergent | some $x$ works in the first row | $\exists x\ \forall \epsilon\ \exists M\ \forall n \ge M$ |

[^lebl-uniq]: Lebl, §2.1, Proposition 2.1.6.
[^lebl-bdd]: Lebl, §2.1, Proposition 2.1.7.
[^lebl-tail]: Lebl, §2.1, Proposition 2.1.15.