---
title: Limit Laws and Monotone Convergence
module: Sequences and Series
moduleNumber: 2
lessonNumber: 2
order: 202
summary: >
  Limits commute with sums, products, quotients, roots, and absolute values and
  preserve non-strict inequalities, so a limit can be assembled from the limits
  of its parts without returning to epsilon and M. The squeeze lemma transfers a
  limit through two envelopes; the monotone convergence theorem produces a limit
  from boundedness alone; and the ratio test settles the geometric and factorial
  standard limits.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "§2.2 Facts about limits of sequences"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

The $\epsilon$-$M$ definition proves individual limits but is clumsy for
computation. The algebraic and order properties below assemble the limit of a
complicated expression from the limits of its parts, and the monotone
convergence theorem produces a limit from boundedness without naming it in
advance. Together they let most limits be computed by inspection.

## Limits and inequalities

The first tool compares an unknown sequence to two known ones that close in on
a common value.

> **Lemma (Squeeze lemma).** Let $\{a_n\}$, $\{x_n\}$, $\{b_n\}$ satisfy
> $a_n \le x_n \le b_n$ for all $n$. If $\{a_n\}$ and $\{b_n\}$ converge with
> $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = x$, then $\{x_n\}$ converges
> and $\lim_{n\to\infty} x_n = x$.

> **Proof.** Given $\epsilon > 0$, choose $M_1$ with $|a_n - x| < \epsilon$ for
> $n \ge M_1$ and $M_2$ with $|b_n - x| < \epsilon$ for $n \ge M_2$. For
> $n \ge M = \max\{M_1, M_2\}$ we get $x - \epsilon < a_n$ and $b_n < x +
> \epsilon$, hence
> $$
> x - \epsilon < a_n \le x_n \le b_n < x + \epsilon,
> $$
> which is $|x_n - x| < \epsilon$.[^lebl-sq] $\square$

$$
% caption: The squeeze lemma. Two envelopes a_n and b_n close on the common
% value L; the trapped sequence x_n is forced to the same limit.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.6,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,3.7);
\draw[acc, thick] (0,2.0) -- (9.6,2.0) node[right, black] {$L$};
% b_n points (upper) and a_n points (lower)
\foreach \n/\yb/\ya/\yx in {1/3.4/0.6/2.7, 2/2.7/1.3/1.5, 3/2.47/1.53/2.3, 4/2.35/1.65/1.8, 5/2.28/1.72/2.15, 6/2.23/1.77/1.9, 7/2.2/1.8/2.08, 8/2.175/1.825/1.95} {
  \fill[black] (\n*1.1,\yb) circle (1.6pt);
  \fill[black] (\n*1.1,\ya) circle (1.6pt);
  \fill[acc] (\n*1.1,\yx) circle (2.2pt);
}
\draw[black, dashed] (1.1,3.4) -- (2.2,2.7) -- (3.3,2.47) -- (4.4,2.35) -- (5.5,2.28) -- (6.6,2.23) -- (7.7,2.2) -- (8.8,2.175);
\draw[black, dashed] (1.1,0.6) -- (2.2,1.3) -- (3.3,1.53) -- (4.4,1.65) -- (5.5,1.72) -- (6.6,1.77) -- (7.7,1.8) -- (8.8,1.825);
\node[black, anchor=west] at (8.9,2.6) {$b_n$};
\node[black, anchor=west] at (8.9,1.4) {$a_n$};
\node[acc, anchor=south] at (1.1,2.75) {$x_n$};
\end{tikzpicture}
$$

> **Worked example.** To find $\lim_{n\to\infty} \frac{1}{n\sqrt{n}}$, note
> $\sqrt{n} \ge 1$ gives $0 \le \frac{1}{n\sqrt{n}} \le \frac{1}{n}$. The
> constant sequence $0$ and $\{1/n\}$ both converge to $0$, so the squeeze lemma
> forces $\frac{1}{n\sqrt{n}} \to 0$.

Limits respect non-strict inequalities, but they can collapse strict ones.

> **Lemma (Order and limits).** If $\{x_n\}$ and $\{y_n\}$ converge and
> $x_n \le y_n$ for all $n$, then $\lim_{n\to\infty} x_n \le \lim_{n\to\infty}
> y_n$.

The strict version is false. With $x_n = -1/n$ and $y_n = 1/n$ we have
$x_n < y_n$ for every $n$, yet both limits equal $0$. Strict inequalities may
weaken to non-strict ones under a limit; this is a common source of error.[^lebl-ord]

## The algebra of limits

Limits pass through the four arithmetic operations.

> **Proposition (Limit laws).** Let $x_n \to x$ and $y_n \to y$. Then
> - $x_n + y_n \to x + y$ and $x_n - y_n \to x - y$;
> - $x_n y_n \to xy$;
> - if $y \ne 0$ and $y_n \ne 0$ for all $n$, then $x_n / y_n \to x/y$.

> **Proof.** _Sum._ Given $\epsilon > 0$, choose $M_1, M_2$ so that
> $|x_n - x| < \epsilon/2$ and $|y_n - y| < \epsilon/2$ past each. For
> $n \ge \max\{M_1, M_2\}$,
> $$
> |(x_n + y_n) - (x + y)| \le |x_n - x| + |y_n - y| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon.
> $$
>
> _Product._ Write $z = xy$ and expand around the limits:
> $$
> x_n y_n - xy = (x_n - x)\,y + x\,(y_n - y) + (x_n - x)(y_n - y).
> $$
> With $K = \max\{|x|, |y|, 1\}$ and each of $|x_n - x|, |y_n - y|$ eventually
> below $\frac{\epsilon}{3K}$, the three terms are each at most $\epsilon/3$,
> giving $|x_n y_n - xy| < \epsilon$. The quotient reduces to the product once
> $1/y_n \to 1/y$ is established, which uses the eventual bound $|y_n| > |y|/2$
> to keep the denominator away from zero.[^lebl-alg] $\square$

Constant sequences turn these into the familiar corollaries: for $c \in
\mathbb{R}$, $\lim(c\,x_n) = c \lim x_n$ and $\lim(c + x_n) = c + \lim x_n$. By
induction, $\lim x_n^k = (\lim x_n)^k$ for every $k \in \mathbb{N}$. Two further
operations commute with limits.

> **Proposition (Roots and absolute value).** If $x_n \to x$ with $x_n \ge 0$
> for all $n$, then $\sqrt{x_n} \to \sqrt{x}$. For any convergent $\{x_n\}$,
> $|x_n| \to |x|$.

The absolute-value statement is immediate from the reverse triangle inequality
$\bigl||x_n| - |x|\bigr| \le |x_n - x|$: the left side is small whenever the
right side is. Its converse fails — $|(-1)^n| \to 1$ while $\{(-1)^n\}$
diverges — so taking absolute values can create convergence that was not
there.[^lebl-root]

$$
% caption: The limit laws compose. Given the limits of two input sequences, the
% limit of any arithmetic combination is the same combination of the limits.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=20mm, minimum height=9mm, align=center, font=\footnotesize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (xn) at (0,1.1) {$x_n$};
\node[box] (yn) at (0,-1.1) {$y_n$};
\node[box, draw=acc, text=acc] (op) at (3.6,0) {sum, product,\\quotient};
\node[box] (lim) at (7.4,0) {same law on\\$x$ and $y$};
\draw[->, acc, thick] (xn) -- (op);
\draw[->, acc, thick] (yn) -- (op);
\draw[->, acc, thick] (op) -- (lim);
\node[black, anchor=south, font=\scriptsize] at (0,1.65) {limit $x$};
\node[black, anchor=north, font=\scriptsize] at (0,-1.65) {limit $y$};
\end{tikzpicture}
$$

One caution: the laws require the pieces to converge first. The limit
$\lim_{n\to\infty}\bigl(\frac{n^2}{n+1} - n\bigr) = -1$ is correct only after
combining over a common denominator; splitting it as
$\lim \frac{n^2}{n+1} - \lim n$ is meaningless, since neither piece converges.

## A convergence test from the definition

Convergence of $\{x_n\}$ to $x$ is the same as convergence of $|x_n - x|$ to
$0$, which gives a reusable test.

> **Proposition (Comparison to a null sequence).** Suppose there is $x \in
> \mathbb{R}$ and a sequence $a_n \to 0$ with $|x_n - x| \le a_n$ for all $n$.
> Then $x_n \to x$.

> **Proof.** Given $\epsilon > 0$, pick $M$ with $a_n = |a_n - 0| < \epsilon$ for
> $n \ge M$. Then $|x_n - x| \le a_n < \epsilon$.[^lebl-null] $\square$

Studying whether $\{x_n\}$ has limit $x$ thus reduces to studying whether the
error sequence $|x_n - x|$ goes to zero, which is often a simpler question.

## The monotone convergence theorem

The limit laws assume a limit and compute it. Monotone convergence produces a
limit from a structural hypothesis, without any candidate value in hand, and it
is the first result here to draw on the
[least-upper-bound property](/real-analysis/foundations/ordered-fields-completeness)
of $\mathbb{R}$.

> **Definition (Monotone).** A sequence is **monotone increasing** if
> $x_n \le x_{n+1}$ for all $n$, and **monotone decreasing** if $x_n \ge
> x_{n+1}$ for all $n$. A sequence that is either is **monotone**.

> **Theorem (Monotone convergence).** A monotone sequence converges if and only
> if it is bounded. If $\{x_n\}$ is monotone increasing and bounded, then
> $$
> \lim_{n\to\infty} x_n = \sup\{x_n : n \in \mathbb{N}\};
> $$
> if monotone decreasing and bounded, the limit is the infimum.

> **Proof (increasing case).** A convergent sequence is bounded, giving one
> direction. Conversely, suppose $\{x_n\}$ is increasing and bounded, and set
> $x = \sup\{x_n : n \in \mathbb{N}\}$, which exists by completeness. Fix
> $\epsilon > 0$. Since $x - \epsilon$ is not an upper bound, some $x_M > x -
> \epsilon$. Because the sequence increases, $x_n \ge x_M > x - \epsilon$ for all
> $n \ge M$, and $x_n \le x$ always, so
> $$
> |x_n - x| = x - x_n \le x - x_M < \epsilon
> $$
> for all $n \ge M$.[^lebl-mct] $\square$

An increasing sequence is automatically bounded below by its first term, so only
an upper bound needs checking; symmetrically for decreasing sequences. The value
of the theorem is that it certifies a limit exists even when computing the
supremum is hard.

$$
% caption: A bounded monotone increasing sequence. Terms climb but never exceed
% the supremum, and they close on it: past any band below the sup, all terms enter.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.6,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,3.5);
\draw[acc, thick] (0,3.0) -- (9.6,3.0) node[right, black] {supremum};
\draw[acc!70, dashed] (0,2.7) -- (9.6,2.7);
\foreach \n/\y in {1/0.8, 2/1.5, 3/2.0, 4/2.35, 5/2.58, 6/2.73, 7/2.82, 8/2.88, 9/2.92}
  {\fill[acc] (\n*1.0,\y) circle (2.2pt); \draw[black, thin] (\n*1.0,0) -- (\n*1.0,\y);}
\end{tikzpicture}
$$

> **Worked example.** The sequence $\{1/\sqrt{n}\}$ is decreasing and bounded
> below by $0$, so it converges. Since $\sqrt{n+1} \ge \sqrt{n}$ gives
> $\frac{1}{\sqrt{n+1}} \le \frac{1}{\sqrt{n}}$, monotone convergence identifies
> the limit as $\inf\{1/\sqrt{n} : n \in \mathbb{N}\}$. Any lower bound $b \ge 0$
> satisfies $b \le 1/\sqrt{n}$, hence $b^2 \le 1/n$ for all $n$; by the
> Archimedean property $b^2 \le 0$, so $b = 0$. The infimum is $0$, and
> $\lim_{n\to\infty} 1/\sqrt{n} = 0$.

$$
% caption: A bounded monotone decreasing sequence falls toward its infimum. Past
% any band above the inf, all terms enter, mirroring the increasing case.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.6,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,3.5);
\draw[acc, thick] (0,0.5) -- (9.6,0.5) node[right, black] {inf};
\draw[acc!70, dashed] (0,0.85) -- (9.6,0.85);
\foreach \n/\y in {1/3.2, 2/2.3, 3/1.75, 4/1.4, 5/1.15, 6/0.98, 7/0.86, 8/0.78, 9/0.72}
  {\fill[acc] (\n*1.0,\y) circle (2.2pt); \draw[black, thin] (\n*1.0,0) -- (\n*1.0,\y);}
\end{tikzpicture}
$$

> **Worked example.** Let $x_1 = 2$ and $x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}$,
> which simplifies to $x_{n+1} = \frac{x_n^2 + 2}{2x_n}$. Induction shows every
> $x_n > 0$ (so the sequence is well defined and bounded below) and that
> $x_n^2 \ge 2$, whence $x_{n+1} \le x_n$: the sequence is decreasing and bounded
> below, so it converges. Writing $x = \lim x_n$ and passing to the limit in
> $2 x_n x_{n+1} = x_n^2 + 2$ gives $2x^2 = x^2 + 2$, so $x^2 = 2$ and, since
> $x \ge 0$, $x = \sqrt{2}$. This is Newton's method for $\sqrt 2$, and it
> converges rapidly.

The order of the argument is mandatory: establish convergence first, then solve
for the limit. Assuming a limit blindly can produce a wrong answer for a
divergent sequence.

## Standard limits and the ratio test

A few limits recur so often they are worth settling once. Each follows from the
behavior of geometric powers.

> **Proposition.** Let $c > 0$. If $c < 1$ then $c^n \to 0$; if $c > 1$ then
> $\{c^n\}$ is unbounded.

> **Proof.** For $c < 1$ the sequence $\{c^n\}$ is positive and decreasing, hence
> convergent; passing to the limit in $c^{n+1} = c \cdot c^n$ gives $x = cx$, so
> $x = 0$. For $c > 1$, apply the first part to $1/c < 1$: given any $B > 0$,
> eventually $(1/c)^n < 1/B$, i.e. $c^n > B$.[^lebl-geo] $\square$

The ratio of consecutive terms of $\{c^n\}$ is exactly $c$. Generalizing to any
sequence whose consecutive ratios tend to a limit gives a broadly applicable
test.

> **Lemma (Ratio test for sequences).** Let $x_n \ne 0$ for all $n$ and suppose
> $L = \lim_{n\to\infty} \frac{|x_{n+1}|}{|x_n|}$ exists. If $L < 1$ then
> $x_n \to 0$; if $L > 1$ then $\{x_n\}$ is unbounded.

When $L = 1$ the test gives no information: the four sequences $1/n \to 0$, the constant
$1 \to 1$, $(-1)^n$ divergent, and $n$ unbounded all have $L = 1$ yet behave
differently. The proof compares $\{x_n\}$ to a geometric sequence $\{r^n\}$ with
$L < r < 1$: past some $M$ the ratios stay below $r$, so $|x_n| \le |x_M|
r^{-M} r^n$, and the right side goes to zero.[^lebl-ratio]

$$
% caption: The ratio test for sequences, read off the limit L of consecutive
% ratios. The boundary case L equal to one carries no information.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=30mm, minimum height=10mm, align=center, font=\footnotesize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc, text=acc] (r) at (0,0) {ratio limit $L$};
\node[box] (lt) at (5.0,1.7) {$L < 1$: terms to $0$};
\node[box] (eq) at (5.0,0) {$L = 1$: no conclusion};
\node[box] (gt) at (5.0,-1.7) {$L > 1$: unbounded};
\draw[->, acc, thick] (r) -- (lt);
\draw[->, black, thick] (r) -- (eq);
\draw[->, acc, thick] (r) -- (gt);
\end{tikzpicture}
$$

Two consequences of the ratio test, used throughout the series lessons:

- $\displaystyle \lim_{n\to\infty} \frac{2^n}{n!} = 0$, since the ratio of consecutive terms is $\frac{2}{n+1} \to 0 < 1$.
- $\displaystyle \lim_{n\to\infty} n^{1/n} = 1$. Given $\epsilon > 0$, the ratio test applied to $\frac{n}{(1+\epsilon)^n}$ shows that sequence tends to $0$, so eventually $n < (1+\epsilon)^n$, i.e. $n^{1/n} < 1 + \epsilon$; and $n^{1/n} \ge 1$ always.

| Sequence | Behavior | Reason |
| --- | --- | --- |
| $c^n$, $0 < c < 1$ | $\to 0$ | geometric decay |
| $c^n$, $c > 1$ | unbounded | reciprocal of the above |
| $2^n / n!$ | $\to 0$ | ratio $\tfrac{2}{n+1} \to 0$ |
| $n^{1/n}$ | $\to 1$ | ratio test on $n/(1+\epsilon)^n$ |
| $1/n^p$, $p > 0$ | $\to 0$ | squeeze / powers of $1/n$ |

## Summary

The limit laws let a limit be computed by decomposing an expression into pieces
with known limits, provided each piece converges. The squeeze lemma and the
null-sequence comparison handle expressions too tangled to decompose. The
monotone convergence theorem, which rests on completeness, gives a limit from
boundedness alone, and the ratio test settles the geometric and factorial
sequences that dominate the study of series.

[^lebl-sq]: Lebl, §2.2, Lemma 2.2.1.
[^lebl-ord]: Lebl, §2.2, Lemma 2.2.3 and the remark following Corollary 2.2.4.
[^lebl-alg]: Lebl, §2.2, Proposition 2.2.5.
[^lebl-root]: Lebl, §2.2, Propositions 2.2.6 and 2.2.7.
[^lebl-null]: Lebl, §2.2, Proposition 2.2.10.
[^lebl-mct]: Lebl, §2.1, Theorem 2.1.10.
[^lebl-geo]: Lebl, §2.2, Proposition 2.2.11.
[^lebl-ratio]: Lebl, §2.2, Lemma 2.2.12.