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 When a sequence cannot be decomposed into arithmetic operations on simpler sequences, a comparison argument is often effective. The squeeze theorem is the principal tool. [theorem:Squeeze Theorem] Let $(a_n)$, $(b_n)$, and $(c_n)$ be real sequences satisfying $a_n \leq b_n \leq c_n$ for all $n \geq N_0$, for some $N_0 \in \mathbb{N}$. If $a_n \to L$ and $c_n \to L$ for some $L \in \mathbb{R}$, then $b_n \to L$. [/theorem] [proof] Let $\varepsilon > 0$. Since $a_n \to L$, there exists $N_1 \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ for all $n \geq N_1$, which gives $L - \varepsilon < a_n$. Since $c_n \to L$, there exists $N_2 \in \mathbb{N}$ such that $|c_n - L| < \varepsilon$ for all $n \geq N_2$, which gives $c_n < L + \varepsilon$. Set $N = \max(N_0, N_1, N_2)$. For all $n \geq N$, \begin{align*} L - \varepsilon < a_n \leq b_n \leq c_n < L + \varepsilon, \end{align*} so $|b_n - L| < \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $b_n \to L$. [/proof] The squeeze theorem is indispensable for sequences involving oscillatory factors. For instance, since $-1 \leq \sin n \leq 1$ for all $n$, we have \begin{align*} -\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}, \end{align*} and squeezing between $-1/n \to 0$ and $1/n \to 0$ gives $(\sin n)/n \to 0$. No algebra-of-limits decomposition works here, because $\sin n$ itself diverges (it oscillates without settling). [example:Nth Root Of N] **Claim.** $\lim_{n \to \infty} n^{1/n} = 1$. **Proof.** For $n \geq 1$, write $n^{1/n} = 1 + h_n$ where $h_n \geq 0$ (since $n^{1/n} \geq 1$). Raising both sides to the $n$-th power, \begin{align*} n = (1 + h_n)^n \geq 1 + \frac{n(n-1)}{2} h_n^2, \end{align*} where we used the binomial theorem and kept only the $k = 0$ and $k = 2$ terms (the $k = 2$ term exists for $n \geq 2$). Rearranging for $n \geq 2$: \begin{align*} h_n^2 \leq \frac{2(n - 1)}{n(n - 1)} = \frac{2}{n}. \end{align*} Therefore $0 \leq h_n \leq \sqrt{2/n}$, and since $\sqrt{2/n} \to 0$, the squeeze theorem gives $h_n \to 0$. Hence $n^{1/n} = 1 + h_n \to 1$. The key step was choosing the $k = 2$ binomial term. The $k = 1$ term yields only $h_n \leq (n-1)/n$, which does not tend to $0$ fast enough. This illustrates a general principle: when bounding $(1 + h)^n$ from below, higher-order binomial terms give stronger estimates. [/example] ## Monotone Convergence and Completeness The algebra of limits and the squeeze theorem share a common prerequisite: one must already know (or suspect) the value of the limit. In many situations --- recursive sequences, infinite products, sequences defined implicitly --- the limit is unknown. The [Monotone Convergence Theorem](/theorems/509) provides an existence result: under the right structural conditions, convergence is guaranteed, and the limit can be determined after the fact. [definition:Bounded Sequence] A sequence $(a_n)_{n \in \mathbb{N}}$ of real numbers is **bounded above** if there exists $M \in \mathbb{R}$ with $a_n \leq M$ for all $n$, **bounded below** if there exists $m \in \mathbb{R}$ with $a_n \geq m$ for all $n$, and **bounded** if it is both bounded above and bounded below, equivalently, if there exists $C > 0$ with $|a_n| \leq C$ for all $n$. [/definition] [definition:Monotone Sequence] A sequence $(a_n)_{n \in \mathbb{N}}$ of real numbers is **increasing** (or **non-decreasing**) if $a_n \leq a_{n+1}$ for all $n$, and **strictly increasing** if $a_n < a_{n+1}$ for all $n$. It is **decreasing** (or **non-increasing**) if $a_n \geq a_{n+1}$ for all $n$, and **strictly decreasing** if $a_n > a_{n+1}$ for all $n$. A sequence is **monotone** if it is either increasing or decreasing. [/definition] [theorem:Monotone Convergence Theorem] Every bounded, monotone sequence of real numbers converges. Specifically: (i) If $(a_n)$ is increasing and bounded above, then $a_n \to \sup\{a_n : n \in \mathbb{N}\}$. (ii) If $(a_n)$ is decreasing and bounded below, then $a_n \to \inf\{a_n : n \in \mathbb{N}\}$. [/theorem] [proof] We prove (i); part (ii) follows by applying (i) to $(-a_n)$. Let $s = \sup\{a_n : n \in \mathbb{N}\}$, which exists by the completeness of $\mathbb{R}$ (the set $\{a_n : n \in \mathbb{N}\}$ is non-empty and bounded above). Let $\varepsilon > 0$. Since $s$ is the least upper bound, $s - \varepsilon$ is not an upper bound, so there exists $N \in \mathbb{N}$ with $a_N > s - \varepsilon$. Since $(a_n)$ is increasing, for all $n \geq N$, \begin{align*} s - \varepsilon < a_N \leq a_n \leq s < s + \varepsilon. \end{align*} Therefore $|a_n - s| < \varepsilon$ for all $n \geq N$, so $a_n \to s$. [/proof] The proof pivots on the existence of the supremum. In $\mathbb{Q}$, the sequence of rational numbers $a_1 = 1$, $a_2 = 1.4$, $a_3 = 1.41$, $\ldots$ (decimal truncations of $\sqrt{2}$) is increasing and bounded above by $2$, yet has no rational supremum, and hence does not converge in $\mathbb{Q}$. The Monotone Convergence Theorem is thus a theorem about $\mathbb{R}$, not about ordered fields in general. Its content is equivalent to the completeness axiom: any ordered field in which every bounded monotone sequence converges is necessarily complete. This equivalence reveals the Monotone Convergence Theorem as a restatement of the deepest structural property of the real line. The connection to [Supremum and Infimum](/page/Supremum%20and%20Infimum) is direct: the limit of a bounded increasing sequence is precisely its supremum. This gives a dynamic interpretation of the supremum --- it is the value that the sequence "fills up to" from below. [example:The Sequence Defining Euler Number] **Claim.** The sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ converges. Its limit is denoted $e$. **Proof of convergence.** We show that $(a_n)$ is increasing and bounded above, then invoke the Monotone Convergence Theorem. *Monotonicity.* By the AM-GM inequality applied to the $n+1$ numbers consisting of $n$ copies of $1 + 1/n$ and one copy of $1$: \begin{align*} \frac{n \cdot (1 + 1/n) + 1}{n + 1} = \frac{n + 2}{n+1} = 1 + \frac{1}{n+1} \geq \left(\left(1 + \frac{1}{n}\right)^n \cdot 1\right)^{1/(n+1)}. \end{align*} Raising both sides to the power $n + 1$ gives $\left(1 + \frac{1}{n+1}\right)^{n+1} \geq \left(1 + \frac{1}{n}\right)^n$, so $a_{n+1} \geq a_n$. *Boundedness.* By the binomial theorem, \begin{align*} a_n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^{n} \frac{1}{k!} \cdot \prod_{j=0}^{k-1}\frac{n - j}{n}. \end{align*} Since each factor $(n - j)/n \leq 1$, we have $a_n \leq \sum_{k=0}^{n} 1/k!$. Using $k! \geq 2^{k-1}$ for $k \geq 1$, we get \begin{align*} a_n \leq 1 + \sum_{k=1}^{n} \frac{1}{2^{k-1}} \leq 1 + \sum_{k=0}^{\infty} \frac{1}{2^k} = 1 + 2 = 3. \end{align*} So $(a_n)$ is increasing and bounded above by $3$. By the Monotone Convergence Theorem, $(a_n)$ converges. The limit, denoted $e$, satisfies $2 \leq e \leq 3$. (In fact, $e = 2.71828\ldots$) [/example] ## Subsequences and Bolzano-Weierstrass Not every bounded sequence converges --- the sequence $(-1)^n$ is bounded but oscillates between $-1$ and $1$ without settling. However, it contains convergent pieces: the even-indexed terms $a_{2k} = 1$ converge to $1$, and the odd-indexed terms $a_{2k+1} = -1$ converge to $-1$. These are examples of convergent subsequences. [definition:Subsequence] Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers. A **subsequence** of $(a_n)$ is a sequence of the form $(a_{n_k})_{k \in \mathbb{N}}$, where $n_1 < n_2 < n_3 < \cdots$ is a strictly increasing sequence of natural numbers. The [function](/page/Function) $k \mapsto n_k$ selects terms from the original sequence while preserving their order and skipping to ever-later positions. [/definition] Part (ii) of the algebra of limits states that if $a_n \to L$, then every subsequence $a_{n_k} \to L$ as well. The contrapositive is a powerful divergence test: if two subsequences converge to different limits, the original sequence diverges. For $(-1)^n$, the subsequences $(a_{2k})$ and $(a_{2k+1})$ converge to $1$ and $-1$ respectively, immediately proving divergence of $(-1)^n$. The remarkable content of the Bolzano-Weierstrass theorem is that divergence of a bounded sequence is always of this mild oscillatory type --- some convergent subsequence can always be extracted. [quotetheorem:171] The Bolzano-Weierstrass theorem is a compactness result: it asserts that bounded subsets of $\mathbb{R}$ are sequentially compact, meaning every sequence in such a set has a convergent subsequence with limit in the closure. The standard proof uses the bisection method: given a bounded sequence in $[a, b]$, bisect the interval, choose the half containing infinitely many terms, and repeat. The nested intervals $[a_k, b_k]$ with $b_k - a_k = (b - a)/2^k$ shrink to a single point $L$ by completeness, and a subsequence can be chosen with one term from each interval, converging to $L$. Boundedness is indispensable. The sequence $a_n = n$ is unbounded, and no subsequence converges: any subsequence $(a_{n_k})$ satisfies $a_{n_k} \geq n_k \geq k$, so $a_{n_k} \to \infty$. The Bolzano-Weierstrass theorem also fails in $\mathbb{Q}$. Consider an enumeration $(q_n)$ of the rationals in $[0, 1]$. This is a bounded sequence in $\mathbb{Q}$, and while subsequences exist that converge in $\mathbb{R}$, those converging to irrational limits (which is "most" of them) do not converge in $\mathbb{Q}$. The theorem's proof requires the completeness of $\mathbb{R}$ at the step where the nested intervals are guaranteed to contain a real number. ### Limit Superior and Limit Inferior The Bolzano-Weierstrass theorem guarantees that a bounded sequence has at least one subsequential limit. In general, there may be many. The limit superior and limit inferior single out the extreme ones. [definition:Limit Superior And Limit Inferior] Let $(a_n)_{n \in \mathbb{N}}$ be a bounded real sequence. The **limit superior** (or **upper limit**) is \begin{align*} \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k = \inf_{n \geq 1} \sup_{k \geq n} a_k. \end{align*} The **limit inferior** (or **lower limit**) is \begin{align*} \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k = \sup_{n \geq 1} \inf_{k \geq n} a_k. \end{align*} [/definition] The sequence $s_n = \sup_{k \geq n} a_k$ is decreasing (taking a supremum over a smaller set can only decrease the value) and bounded below (since $(a_n)$ is bounded), so $s_n$ converges by the Monotone Convergence Theorem --- this is why the limit in the definition exists. Similarly for $\inf_{k \geq n} a_k$. A bounded sequence converges if and only if $\limsup a_n = \liminf a_n$, in which case the common value is the limit. For $(-1)^n$, we have $\limsup = 1$ and $\liminf = -1$, confirming divergence and identifying the two extreme subsequential limits. [example:Subsequences Of A Bounded Divergent Sequence] Consider $a_n = (-1)^n (1 - 1/n)$. The terms alternate in sign and approach $\pm 1$: \begin{align*} a_1 = 0, \quad a_2 = \tfrac{1}{2}, \quad a_3 = -\tfrac{2}{3}, \quad a_4 = \tfrac{3}{4}, \quad a_5 = -\tfrac{4}{5}, \quad \ldots \end{align*} The even-indexed subsequence is $a_{2k} = 1 - 1/(2k) \to 1$, and the odd-indexed subsequence is $a_{2k+1} = -(1 - 1/(2k+1)) \to -1$. Therefore $\limsup a_n = 1$ and $\liminf a_n = -1$. Since $\limsup \neq \liminf$, the sequence diverges, but Bolzano-Weierstrass guarantees (and we have explicitly found) convergent subsequences. Every subsequential limit lies in $[-1, 1]$, and $\pm 1$ are achieved. [/example] ## Cauchy Sequences and the Completeness of R Every convergence theorem presented so far requires some external information: the algebra of limits requires knowing the limit, monotone convergence requires knowing the sequence is monotone and bounded, Bolzano-Weierstrass guarantees only a subsequence. The Cauchy criterion provides a characterisation of convergence that is entirely intrinsic --- it depends only on the mutual distances between terms, not on any candidate limit. The idea is that if a sequence converges, its terms must eventually be close to each other, because they are all close to the limit. The Cauchy condition captures this "internal clustering" and, in $\mathbb{R}$, it turns out to be sufficient as well as necessary. We refer to [Cauchy Sequence](/page/Cauchy%20Sequence) for the formal definition: a sequence $(a_n)$ is Cauchy if for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $|a_m - a_n| < \varepsilon$ for all $m, n \geq N$. [quotetheorem:172] The forward direction (convergent implies Cauchy) holds in any metric space and follows from the triangle inequality: if $a_n \to L$, then for $n, m \geq N$, \begin{align*} |a_m - a_n| \leq |a_m - L| + |L - a_n| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} The reverse direction (Cauchy implies convergent) is the deep content: it is equivalent to the completeness of $\mathbb{R}$. The proof proceeds in two steps. First, every Cauchy sequence is bounded (choose $N$ for $\varepsilon = 1$; then $|a_n| \leq \max(|a_1|, \ldots, |a_{N-1}|, |a_N| + 1)$ for all $n$). Second, by Bolzano-Weierstrass, the bounded sequence has a convergent subsequence $a_{n_k} \to L$. Finally, the Cauchy condition forces the full sequence to converge to the same $L$: given $\varepsilon > 0$, choose $K$ large enough that both $|a_m - a_n| < \varepsilon/2$ for $m, n \geq N$ and $|a_{n_k} - L| < \varepsilon/2$ for $k \geq K$, then for $n \geq \max(N, n_K)$, \begin{align*} |a_n - L| \leq |a_n - a_{n_K}| + |a_{n_K} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} The power of the Cauchy criterion is that it certifies convergence without requiring knowledge of the limit. This is essential in situations where the limit is difficult to identify explicitly --- for instance, when proving the convergence of a [Series](/page/Series) by showing its partial sums form a Cauchy sequence. It also provides the template for completeness in general [metric spaces](/page/Metric%20Space): a metric space is called **complete** if every Cauchy sequence in it converges. By the theorem above, $\mathbb{R}$ is complete; $\mathbb{Q}$ is not. [example:A Cauchy Sequence Whose Limit Is Not Obvious] Consider the sequence of partial sums $s_n = \sum_{k=1}^{n} 1/k^2$. We show $(s_n)$ is Cauchy without computing its limit. For $m > n$, \begin{align*} |s_m - s_n| = \sum_{k=n+1}^{m} \frac{1}{k^2} \leq \sum_{k=n+1}^{m} \frac{1}{k(k-1)} = \sum_{k=n+1}^{m} \left(\frac{1}{k-1} - \frac{1}{k}\right) = \frac{1}{n} - \frac{1}{m} < \frac{1}{n}. \end{align*} Given $\varepsilon > 0$, choose $N > 1/\varepsilon$. Then for all $m > n \geq N$, $|s_m - s_n| < 1/n \leq 1/N < \varepsilon$. By the Cauchy criterion, $(s_n)$ converges. The limit turns out to be $\pi^2/6$ (Euler's Basel solution), but the Cauchy criterion establishes convergence without this remarkable identity. The comparison $1/k^2 \leq 1/(k(k-1))$ and the resulting telescoping sum are the essential ingredients. [/example] ## Rates and Modes of Convergence Knowing that a sequence converges is the first step; understanding how fast it converges is often equally important, both for theoretical insight and for computational practice. ### Eventually vs. Frequently The $\varepsilon$-$N$ definition requires that $|a_n - L| < \varepsilon$ for all $n \geq N$ --- the sequence is **eventually** in the $\varepsilon$-band. A weaker condition asks only that the sequence **frequently** enters the band, meaning for every $N$ there exists $n \geq N$ with $|a_n - L| < \varepsilon$. This is the difference between convergence and having $L$ as a subsequential limit. The sequence $(-1)^n$ is frequently within $\varepsilon$ of $1$ (take even-indexed terms) but not eventually so. ### Fast and Slow Convergence Among convergent sequences, the rate at which the error $|a_n - L|$ decays varies enormously. A sequence converges **geometrically** (or exponentially) if there exist $C > 0$ and $0 < r < 1$ such that $|a_n - L| \leq Cr^n$. The Babylonian square-root sequence from the introduction converges even faster --- its error satisfies $|x_{n+1} - \sqrt{2}| \leq C|x_n - \sqrt{2}|^2$, so the number of correct digits roughly doubles at each step (quadratic convergence). At the other extreme, $a_n = 1/n$ converges to $0$, but $|a_n - 0| = 1/n$ decreases only polynomially. The sequence $a_n = 1/\log n$ (for $n \geq 2$) converges even more slowly: to achieve $|a_n| < 0.01$ requires $n > e^{100}$, an astronomically large index. Such sequences are theoretically convergent but computationally useless for approximation. ### Divergence to Infinity A sequence $(a_n)$ **diverges to $+\infty$** if for every $M \in \mathbb{R}$, there exists $N \in \mathbb{N}$ such that $a_n > M$ for all $n \geq N$. We write $a_n \to +\infty$. This is not convergence (there is no real number serving as a limit), but it is a definite mode of divergence, qualitatively different from the oscillatory divergence of $(-1)^n$. The sequence $a_n = n^2$ diverges to $+\infty$; the sequence $a_n = (-1)^n n$ diverges but not to $+\infty$ or $-\infty$. [example:Comparing Convergence Rates] Consider the three sequences $a_n = 1/n$, $b_n = 1/2^n$, and $c_n = 1/\log(n+1)$, all converging to $0$. To achieve accuracy $\varepsilon = 10^{-6}$: - For $a_n = 1/n$: need $n > 10^6$, so $N = 10^6 + 1$ suffices. - For $b_n = 1/2^n$: need $2^n > 10^6$, so $n > 6\log_{2} 10 \approx 19.9$, giving $N = 20$. - For $c_n = 1/\log(n+1)$: need $\log(n+1) > 10^6$, so $n > e^{10^6} - 1$, a number with over $400{,}000$ digits. The geometric sequence $b_n$ reaches any prescribed accuracy in roughly $\log(1/\varepsilon)$ steps. The polynomial sequence $a_n$ requires $1/\varepsilon$ steps. The logarithmic sequence $c_n$ requires $e^{1/\varepsilon}$ steps --- exponentially many. This hierarchy --- geometric, polynomial, logarithmic --- is fundamental throughout analysis and computation. [/example] ## References - W. Rudin, *Principles of Mathematical Analysis*, 3rd edition, McGraw-Hill, 1976. - T. Tao, *Analysis I*, 3rd edition, Springer, 2016. - T. W. K&ouml;rner, *A Companion to Analysis*, AMS, 2004.