The ancient Babylonians discovered that starting from any reasonable guess $x_1 > 0$ for the square root of $2$, the recursive rule

\begin{align*} x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \end{align*}

produces successive approximations that stabilise with astonishing speed. Beginning with $x_1 = 1$, one obtains $x_2 = 3/2$, $x_3 = 17/12$, $x_4 = 577/408$ --- already accurate to five decimal places. Every term is rational, yet the sequence homes in on $\sqrt{2}$, a number that is not rational at all. Two fundamental questions emerge. First, what does it mean, precisely, to say that a sequence of numbers "approaches" a value it never reaches? Second, why should such a limit exist when the target lies outside the number system in which we are computing? The theory of convergence answers both questions: the $\varepsilon$-$N$ definition captures the first, and the completeness of $\mathbb{R}$ resolves the second.

[motivation] ### The naive notion: "getting closer" A first attempt at defining convergence might be: a sequence $(a_n)$ converges to $L$ if each term is closer to $L$ than the previous one, that is, $|a_{n+1} - L| < |a_n - L|$ for all $n$. This captures something real about many convergent [sequences](/page/Sequence), but it is far too restrictive and simultaneously too permissive. It is too restrictive because perfectly well-behaved sequences violate it. Define $a_n = 1/n + (-1)^n / n^2$. Then $a_n \to 0$, yet the terms do not decrease monotonically in distance from $0$ --- the odd-indexed terms are slightly closer than the "getting closer" pattern would predict, while the even-indexed terms overshoot slightly, so consecutive distances oscillate. It is too permissive in spirit because "getting closer" does not guarantee arrival. The sequence $a_n = 1 + 1/n$ has $|a_{n+1} - 2| < |a_n - 2|$ for all $n$, so it is "getting closer" to $2$, yet it converges to $1$, not to $2$. Being attracted toward a value is not the same as being eventually trapped near it. ### The need for epsilon-N The correct intuition is not "getting closer" but "eventually staying close, no matter how close you demand." For any tolerance $\varepsilon > 0$, no matter how small, there should exist a stage $N$ beyond which every term $a_n$ satisfies $|a_n - L| < \varepsilon$. The universal quantifier on $\varepsilon$ prevents the sequence from being "close" in only a weak sense, and the existential quantifier on $N$ allows a finite number of initial terms to misbehave. This is the $\varepsilon$-$N$ definition, and its precision is what makes rigorous analysis possible. ### Completeness: why the field matters Even with the correct definition in hand, convergence depends on the ambient number system. Consider the sequence of rational numbers \begin{align*} 1, \; 1.4, \; 1.41, \; 1.414, \; 1.4142, \; \ldots \end{align*} obtained by truncating the decimal expansion of $\sqrt{2}$. Each term is rational, and the sequence is Cauchy in $\mathbb{Q}$ --- the terms cluster ever more tightly --- yet $\sqrt{2} \notin \mathbb{Q}$, so the sequence has no limit in $\mathbb{Q}$. In the rational numbers, the "gap" where $\sqrt{2}$ should be is genuinely empty, and the sequence, despite its best efforts, converges to nothing. The real numbers $\mathbb{R}$ are constructed precisely to fill such gaps. The completeness axiom --- equivalently stated as every non-empty [set](/page/Set) bounded above having a supremum, or every Cauchy sequence converging --- ensures that [limits](/page/Limit) "exist when they should." Every major convergence theorem in this article relies, at some crucial point, on completeness. [/motivation]

## Definition

[definition:Convergence Of A Real Sequence] Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and let $L \in \mathbb{R}$. We say that $(a_n)$ **converges** to $L$, and write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$, if \begin{align*} \forall \, \varepsilon > 0, \; \exists \, N \in \mathbb{N} \text{ such that } \forall \, n \geq N, \; |a_n - L| < \varepsilon. \end{align*} A sequence that does not converge is called **divergent**. [/definition]

Each quantifier plays a distinct role. The universal quantifier $\forall \, \varepsilon > 0$ says that no positive tolerance is too stringent: convergence means eventual closeness to any prescribed degree. The existential quantifier $\exists \, N$ says that we are allowed a "warm-up" period; only the tail of the sequence matters. The final universal quantifier $\forall \, n \geq N$ insists that closeness is not a one-off event --- once the sequence enters the $\varepsilon$-band around $L$, it stays there permanently.

The absolute value $|a_n - L|$ measures distance on the real line, so the condition $|a_n - L| < \varepsilon$ says that $a_n$ lies in the open interval $(L - \varepsilon, L + \varepsilon)$. Convergence thus means that every open interval centred at $L$, no matter how narrow, eventually captures the entire tail of the sequence. This geometric viewpoint connects directly to the [topology](/page/Topology) of $\mathbb{R}$: a sequence converges to $L$ if and only if every open neighbourhood of $L$ contains all but finitely many terms of the sequence.

[example:Convergence Of One Over N] **Claim.** The sequence $a_n = 1/n$ converges to $0$. **Proof.** Let $\varepsilon > 0$ be given. By the Archimedean property of $\mathbb{R}$, there exists $N \in \mathbb{N}$ with $N > 1/\varepsilon$. Then for all $n \geq N$, \begin{align*} |a_n - 0| = \frac{1}{n} \leq \frac{1}{N} < \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $a_n \to 0$. This argument exhibits the standard structure of an $\varepsilon$-$N$ proof: receive $\varepsilon$, manufacture $N$ (typically by inverting the bound), and verify the inequality for all $n \geq N$. The Archimedean property, which guarantees that $\mathbb{N}$ is unbounded in $\mathbb{R}$, is the engine that makes $N$ exist. [/example]

[example:Convergence Of Geometric Sequence] **Claim.** For $r \in \mathbb{R}$ with $|r| < 1$, the sequence $a_n = r^n$ converges to $0$. **Proof.** If $r = 0$, every term is $0$ and the result is immediate. Otherwise, write $|r| = 1/(1 + h)$ for some $h > 0$ (possible since $|r| < 1$). By the binomial inequality, $(1 + h)^n \geq 1 + nh$ for all $n \in \mathbb{N}$, so \begin{align*} |a_n - 0| = |r|^n = \frac{1}{(1+h)^n} \leq \frac{1}{1 + nh} < \frac{1}{nh}. \end{align*} Given $\varepsilon > 0$, choose $N \in \mathbb{N}$ with $N > 1/(h\varepsilon)$. Then for all $n \geq N$, \begin{align*} |r^n| < \frac{1}{nh} \leq \frac{1}{Nh} < \varepsilon. \end{align*} The substitution $|r| = 1/(1+h)$ is a recurring technique: it converts the problem of controlling a geometric decay into the easier problem of controlling a polynomial growth in the denominator. [/example]

## The Algebra of Limits

Computing limits directly from the $\varepsilon$-$N$ definition for every sequence would be impractical. The algebra of limits provides a toolkit: once a few basic limits are established (such as $1/n \to 0$ and $r^n \to 0$ for $|r| < 1$), more complex limits can be assembled from arithmetic operations on sequences.

[quotetheorem:170]

The [uniqueness of limits](/theorems/625) (part (i)) deserves emphasis. If $a_n \to L$ and $a_n \to M$ with $L \neq M$, then setting $\varepsilon = |L - M|/2$ yields a contradiction: the tail of the sequence cannot simultaneously lie within distance $\varepsilon$ of both $L$ and $M$, since the $\varepsilon$-balls around $L$ and $M$ are disjoint. This argument uses the Hausdorff property of the real line --- distinct points have disjoint open neighbourhoods.

The product rule (part (v)) is subtler than it first appears. The proof splits $|a_n b_n - LM|$ using the identity

\begin{align*} a_n b_n - LM = a_n(b_n - M) + M(a_n - L), \end{align*}

and controls each summand separately. The key auxiliary fact is that every convergent sequence is bounded: if $a_n \to L$, then there exists $C > 0$ with $|a_n| \leq C$ for all $n$. This boundedness is what keeps $|a_n(b_n - M)|$ small --- one factor is bounded while the other tends to zero.

The reciprocal rule (part (vi)) requires $L \neq 0$. This hypothesis is essential, not merely convenient. If $a_n = 1/n$, then $a_n \to 0$, but $1/a_n = n$ diverges to infinity. The proof of the reciprocal rule uses the fact that if $a_n \to L \neq 0$, then $|a_n| \geq |L|/2$ for all sufficiently large $n$, providing a uniform lower bound on the denominators.

[example:Limit Of A Rational Expression] **Claim.** $\displaystyle\lim_{n \to \infty} \frac{3n^2 + 5n}{2n^2 - 1} = \frac{3}{2}$. **Proof.** Divide numerator and denominator by $n^2$: \begin{align*} \frac{3n^2 + 5n}{2n^2 - 1} = \frac{3 + 5/n}{2 - 1/n^2}. \end{align*} Since $1/n \to 0$ (shown above), the constant multiple rule gives $5/n \to 0$, the product rule gives $1/n^2 = (1/n)(1/n) \to 0$, and the sum rule gives \begin{align*} 3 + \frac{5}{n} \to 3, \qquad 2 - \frac{1}{n^2} \to 2. \end{align*} Since the denominator limit $2 \neq 0$, the quotient rule (combining the reciprocal and product rules) yields \begin{align*} \frac{3 + 5/n}{2 - 1/n^2} \to \frac{3}{2}. \end{align*} This "divide by the dominant power" strategy reduces the problem to known limits ($1/n^k \to 0$) and the algebra of limits. [/example]

### The Squeeze Theorem