5.1 Sequences and epsilon-N limits

A sequence may have many terms close to a number and still return repeatedly far away from it. Convergence requires more: for every prescribed accuracy, all sufficiently late terms must remain close. The epsilon-N definition makes that requirement precise.

A sequence is a function, not just a pattern

Many students first meet sequences as lists such as

$$1,2,3,4,\ldots$$

or

$$1,\frac12,\frac13,\frac14,\ldots$$

The familiar idea must be reinterpreted in a more precise way.

Definition 5.1.1

(Sequence in a set)

Let $X$ be a set. A sequence in $X$ is a function

$$N\to X.$$

If the image of $n\in N$ is written $x_n$, then the sequence is denoted $(x_n)$.

So a sequence is not required to come from a simple formula. The only formal requirement is that each natural number $n$ is assigned an element $x_n$ of the target set.

Definition 5.1.2

(Rational and real sequences)

A sequence of rational numbers is a map $N\to Q$. A sequence of real numbers is a map $N\to R$.

This functional viewpoint matters later. It reminds you that a sequence has a domain, a codomain, and an indexing variable. The dots $\ldots$ are only informal shorthand.

A tolerance for the whole tail

For a sequence, the index specifies how far along the list a term lies. A threshold $N$ separates a finite initial segment from the entire remaining tail. To say that the sequence approaches $L$, each positive tolerance must admit a threshold after which every term lies within that tolerance of $L$.

The formal definition

Definition 5.1.3

(Limit of a real sequence)

A sequence $(x_n)$ of real numbers has limit $L\in R$ if for every positive real number $\varepsilon\gt0$, there exists $N\in N$ such that for all $n\gt N$,

$$|L-x_n|\lt\varepsilon.$$

In that case we write

$$L=\lim_{n\to\infty}x_n$$

or say that $(x_n)$ converges to $L$.

The order of quantifiers makes the permitted dependence of each choice explicit:

$$\lim_{n\to\infty}x_n=L \iff \forall \varepsilon\gt0\ \exists N\in N\ \forall n\gt N,\ |x_n-L|\lt\varepsilon.$$

How to read the quantifiers correctly

The definition is easier once you separate its jobs.

1. $\forall \varepsilon\gt0$: you do not get to choose only one error bar. The sequence must eventually fit inside every positive tolerance band.
2. $\exists N\in N$: after seeing the chosen $\varepsilon$, you are allowed to pick a tail-starting point $N$.
3. $\forall n\gt N$: once you pass that point, every later term must stay inside the band.

So convergence is an eventual-tail statement. The first few terms may behave badly. That does not matter. What matters is the behavior far enough out in the sequence.

Common mistake

(N may depend on epsilon, but not on n)

When proving convergence, you are allowed to choose $N$ after the tolerance $\varepsilon$ is given. But once $N$ is fixed, the inequality must hold for every $n\gt N$. You are not allowed to choose a different $N$ for each later term.

The strict inequality in the definition is $n\gt N$. If an estimate is proved for $n≥K$, where $K$ is an integer with $K≥1$, choosing $N=K$ is safe because $n\gt N$ implies $n≥K+1≥K$. If the estimate needs $n\gt K$, choose an integer $N$ with $N≥K$; the choice of $N$ is made after $\varepsilon$ but before the arbitrary later index.

First worked example: $1/n \to 0$

A basic example is that

$$\frac11,\frac12,\frac13,\frac14,\ldots$$

has limit $0$.

Worked example 5.1.4

(Proving that $\lim\limits_{n\to\infty} 1/n = 0$)

Here $n≥1$, so the reciprocal is defined even if the course convention includes 0 in $N$. Let $\varepsilon\gt0$ be given. We want $|1/n-0|\lt\varepsilon$, that is,

$$\frac1n\lt\varepsilon.$$

This is guaranteed once

$$n\gt\frac1\varepsilon.$$

The Archimedean property provides a natural number $N$ with $N\gt1/\varepsilon$. Then every $n\gt N$ satisfies

$$\left|\frac1n-0\right|=\frac1n\lt\varepsilon.$$

Therefore

$$\lim_{n\to\infty}\frac1n=0.$$

This proof is typical. Start from the quantity you need to make small, simplify it, and then choose $N$ large enough.

Second worked example: $\frac{5n+2}{3n-7} \to \frac53$

A rational expression requires control of the denominator as well as the error.

Worked example 5.1.5

(Proving that $\lim\limits_{n\to\infty}\frac{5n+2}{3n-7}=\frac53$)

We compute

$$\left|\frac{5n+2}{3n-7}-\frac53\right| = \left|\frac{3(5n+2)-5(3n-7)}{3(3n-7)}\right| = \left|\frac{41}{3(3n-7)}\right|.$$

For large $n$, the denominator is positive, so

$$\left|\frac{41}{3(3n-7)}\right|=\frac{41}{3(3n-7)}.$$

Now pick $N$ large enough that $3n-7\ge 2n$ whenever $n\gt N$. Then

$$\frac{41}{3(3n-7)}\le \frac{41}{6n}.$$

So it is enough to ensure

$$\frac{41}{6n}\lt\varepsilon.$$

Choose $N$ so large that both $3n-7\ge 2n$ and $n\gt41/(6\varepsilon)$ hold for all $n\gt N$. Then

$$\left|\frac{5n+2}{3n-7}-\frac53\right|\lt\varepsilon.$$

Hence

$$\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53.$$

The algebra looks longer, but the logical structure is exactly the same as for $1/n$: rewrite the error term and make it small by forcing $n$ to be large.

Designing an epsilon threshold

The rational example requires two conditions on the same tail: the denominator estimate $3n-7≥2n$ needs $n≥7$, and the error estimate needs $n>41/(6\varepsilon)$. Given $\varepsilon>0$, choose an integer

$$N\ge\max\left(7,\frac{41}{6\varepsilon}\right).$$

Then every $n>N$ satisfies both conditions, so $41/(6n)<\varepsilon$. This is the dependency order of the proof: derive the bound, choose one threshold for all restrictions, then verify it for an arbitrary later index.

The index convention also matters. The expression $1/n$ is defined for $n≥1$. To make it a sequence on $\mathbb N=\{0,1,2,\ldots\}$, specify the value at $n=0$ separately. A finite initial change does not affect convergence.

Worked example 5.1.6

(A finite prefix and an identical tail)

Let $x_0=100$ and $x_n=1/n$ for $n≥1$. Given $\varepsilon>0$, choose an integer $N>1/\varepsilon$ with $N≥1$. Then every $n>N$ satisfies $|x_n|=1/n<\varepsilon$; the exceptional first term lies before the threshold. Thus $x_n→0$.

Constant sequences

For the constant sequence $x_n=0$, the limit is immediate.

Worked example 5.1.7

(Constant sequences converge to their constant value)

Let $x_n=0$ for all $n$. Then for every $\varepsilon\gt0$,

$$|x_n-0|=0\lt\varepsilon$$

for every natural number $n$.

So any natural number can serve as $N$, and

$$\lim_{n\to\infty}x_n=0.$$

What convergence means geometrically

If $\lim\limits_{n\to\infty}x_n=L$, then no matter how narrow a band you draw around $L$, the tail of the sequence eventually stays inside it.

Equivalently:

• you may reject finitely many early terms;
• afterward the sequence cannot keep escaping the chosen $\varepsilon$-band.

That is why convergence is stronger than “having many terms near $L$”. The definition requires all sufficiently late terms to be near $L$.

Figure. Convergence means more than seeing some terms near $L$. After a large enough index $N$, the whole tail must remain inside the chosen $\varepsilon$-band.

Compare tails interactively

In the figure below, choose a sequence and a tolerance $\varepsilon$, then move $N$ to test a proposed tail. The plotted signed errors $x_n-L$ show which terms lie in the open band $-\varepsilon\lt x_n-L\lt\varepsilon$. Check the terms with $n\gt N$, and compare the figure with the accompanying algebraic explanation: a finite plot illustrates the tail, but the estimate must justify every later term.

Proof tools for sequence limits

The definition controls a tail, so several useful consequences follow by making that tail explicit.

Theorem 5.1.8

(Limits are unique)

A real sequence cannot converge to two different limits. Suppose that $x_n→L$ and $x_n→M$ with $L≠M$. Choose

$$\varepsilon=\frac{|L-M|}{3}\gt0.$$

For sufficiently large $n$, both $|x_n-L|\lt\varepsilon$ and $|x_n-M|\lt\varepsilon$. The triangle inequality then gives

$$|L-M|\le |L-x_n|+|x_n-M|\lt2\varepsilon=\frac{2|L-M|}{3},$$

which is impossible. Thus the limit, when it exists, is unique.

Theorem 5.1.9

(Every convergent sequence is bounded)

If $x_n→L$, then there are real numbers $m$ and $M$ such that $m≤x_n≤M$ for every index. Use $\varepsilon=1$ to choose $N$ so that $n\gt N$ implies $L-1\lt x_n\lt L+1$. The terms with $n≤N$ form a finite set, so they have a minimum $m_0$ and maximum $M_0$. Taking

$$m=\min(m_0,L-1),\qquad M=\max(M_0,L+1)$$

bounds both the finite initial segment and the tail.

Theorem 5.1.10

(Absolute values preserve limits)

If $x_n→L$, then $|x_n|→|L|$. The reverse triangle inequality says

$$\bigl||x_n|-|L|\bigr|\le |x_n-L|.$$

Given $\varepsilon\gt0$, use the same index supplied for $|x_n-L|\lt\varepsilon$. The displayed inequality transfers that estimate directly to the absolute values.

Worked example 5.1.11

(Derive both triangle inequalities)

The triangle inequality applied to $(x-y)+(y-z)$ gives

$$|x-z|=|(x-y)+(y-z)|\le |x-y|+|y-z|.$$

For the reverse triangle inequality, apply the triangle inequality to $x=(x-y)+y$ and then rearrange:

$$|x|\le |x-y|+|y|\quad\Longrightarrow\quad |x|-|y|\le |x-y|.$$

Interchanging $x$ and $y$ gives $|y|-|x|≤|x-y|$; combining the two inequalities is exactly $\bigl||x|-|y|\bigr|≤|x-y|$.

Algebra of sequence limits

Once the basic examples are established, the limit laws let us combine them. Their proofs show why the same threshold must control every error.

Theorem 5.1.12

(Sum and product laws for sequences)

If $x_n\to L$ and $y_n\to M$, then $x_n+y_n\to L+M$ and $x_ny_n\to LM$.

Proof. For the sum, make each error smaller than $\varepsilon/2$ and use the triangle inequality on the common tail. For the product, first use $x_n\to L$ with tolerance $1$ to obtain the fixed bound $|x_n|<|L|+1=:A$. Put $B=\max(1,|M|)$. Given $\varepsilon>0$, choose a common tail on which

$$|y_n-M|<\frac{\varepsilon}{2A},\qquad |x_n-L|<\frac{\varepsilon}{2B}.$$

All bounds now hold for the same arbitrary $n$ beyond the threshold, so

$$|x_ny_n-LM| \le |x_n||y_n-M|+|M||x_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

The preliminary bound prevents the factor $x_n$ from magnifying a small error without control; using $B$ also covers $M=0$.

Theorem 5.1.13

(Reciprocals of nonzero limits)

If $x_n\to L$ and $L\ne0$, then $1/x_n\to1/L$ on the tail where the reciprocal is defined.

Proof. Choose a tail with $|x_n-L|<|L|/2$. Then $|x_n|>|L|/2$, and

$$\left|\frac1{x_n}-\frac1L\right| =\frac{|x_n-L|}{|x_n||L|} \le\frac{2|x_n-L|}{|L|^2}.$$

Given $\varepsilon>0$, enlarge the threshold so that $|x_n-L|<\varepsilon|L|^2/2$ also holds. The error is then below $\varepsilon$, and every division is justified by the earlier lower bound.

Worked example 5.1.14

(Apply the sequence laws)

For $n\ge1$, the sum law gives $1+1/n\to1$. The product law then gives $(1+1/n)/n\to0$ by writing the expression as $(1+1/n)(1/n)$. Both conclusions use the proved limit $1/n\to0$ with its stated domain.

Further examples: comparison, oscillation, and growth

Worked example 5.1.15

(Compare two rational-expression limits)

For

$$x_n=\frac{5n^2+n+7}{n^3+3n^2+3n+1},$$

we have, for $n≥1$,

$$0\le x_n\le\frac{13n^2}{n^3}=\frac{13}{n},$$

because the denominator is at least $n^3$ and $5n^2+n+7≤13n^2$. Given $\varepsilon\gt0$, choose an integer $N\gt13/\varepsilon$. Then $n\gt N$ implies $|x_n|\lt\varepsilon$, so $x_n→0$.

Now consider a numerator with an oscillating term:

$$y_n=\frac{n^2-(-1)^n n-13}{n^2+n+1}.$$

Subtracting 1 gives

$$y_n-1=\frac{-(-1)^n n-n-14}{n^2+n+1}.$$

Using $|(-1)^n|=1$, for $n≥1$ we obtain

$$|y_n-1|\le\frac{2n+14}{n^2}\le\frac{16}{n}.$$

Choose $N\gt16/\varepsilon$; then $y_n→1$.

Worked example 5.1.16

(The sequence $2^n$ has no finite limit)

For every $n≥0$, induction gives $2^n≥n+1$: it is true at $n=0$, and $2^{n+1}=2\cdot2^n≥2(n+1)≥n+2$. Thus $2^n$ is unbounded. A convergent real sequence is bounded, so the sequence $2^n$ cannot converge to a real number.

Worked example 5.1.17

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

For $n≥1$,

$$0\lt\sqrt{n+1}-\sqrt n =\frac{1}{\sqrt{n+1}+\sqrt n} \le\frac{1}{\sqrt n}.$$

Given $\varepsilon\gt0$, choose an integer $N\gt1/\varepsilon^2$. Then $n\gt N$ implies $1/\sqrt n\lt\varepsilon$, proving $\sqrt{n+1}-\sqrt n\to0$.

Worked example 5.1.18

(The discrete limit $n^2/4^n$)

For $n≥4$, induction shows $n≤2^{n/2}$. The base case is $4≤4$; if the inequality holds at $n≥4$, then $(n+1)/n≤5/4\lt\sqrt2$, so $n+1≤2^{(n+1)/2}$. Consequently,

$$0\le\frac{n^2}{4^n}\le\frac{2^n}{4^n}=2^{-n}.$$

The earlier induction bound $2^n\ge n+1$ gives $2^{-n}\le1/(n+1)$. Given $\varepsilon\gt0$, choose an integer $N>\max(4,1/\varepsilon)$. Then $n>N$ gives $2^{-n}\le1/(n+1)<\varepsilon$, hence $n^2/4^n\to0$.

Worked example 5.1.19

(Finite changes do not change a sequence limit)

If two sequences agree for all $n\gt N_0$ and one converges to $L$, choose the limit threshold for a given $\varepsilon$ and enlarge it to exceed $N_0$. The two tails then have identical terms and the same estimate, so the other sequence also converges to $L$.

Quick checks

Quick check

(In the definition of $\lim\limits_{n\to\infty}x_n=L$, what is the role of N?)

Answer in terms of the tail of the sequence.

Solution · Answer

$N$ marks the point after which every later term must lie within the chosen $\varepsilon$-band around $L$. It is a tail-starting index.

Quick check

(Why does changing the first ten terms of a convergent sequence never destroy convergence?)

Look at which indices the definition actually controls.

Solution · Answer

Because convergence only asks for a property of all terms with $n\gt N$ for some large $N$. Changing finitely many initial terms does not change the eventual tail behavior.

Exercises

Quick check

(Write the sequence-limit definition entirely in symbols.)

Do not omit the order of the quantifiers.

Solution · Guided solution

The symbolic form is

$$\lim_{n\to\infty}x_n=L \iff \forall \varepsilon\gt0\ \exists N\in N\ \forall n\gt N,\ |x_n-L|\lt\varepsilon.$$

The order matters: first the tolerance is chosen, then $N$, then every later term is checked.

Quick check

(Show directly from the definition that $\lim\limits_{n\to\infty} 1/(2n)=0$.)

Compare it with the proof for $1/n$.

Solution · Guided solution

Let $\varepsilon\gt0$. We want

$$\left|\frac{1}{2n}-0\right|=\frac{1}{2n}\lt\varepsilon.$$

This is true once $n\gt1/(2\varepsilon)$. Choose a natural number $N$ with $N\gt1/(2\varepsilon)$. Then every $n\gt N$ satisfies $1/(2n)\lt\varepsilon$, so $1/(2n)\to 0$.

Quick check

(Why is it not enough for infinitely many terms of a sequence to lie close to L?)

Contrast “infinitely many” with “all sufficiently late”.

Solution · Guided solution

Because convergence requires the whole tail to stay close to $L$, not merely a subcollection of terms. A sequence can visit points near $L$ infinitely often and still keep escaping far away at later indices.

Prerequisites and continuation

Read this after 4.6 Decimal expansions and irrational numbers and 4.3 Completeness and gaps in Q. Then continue to 5.2 Cauchy sequences and another model of the reals.