Sequences: Convergence, Cauchy Sequences, Subsequences

⚡ Key Takeaways

A sequence $(a_n)$ converges to a limit $L$ if its terms get arbitrarily close to $L$ for all large $n$ ; it is Cauchy if its terms get arbitrarily close to each other. In $\mathbb{R}$ (and any complete space) these two notions coincide. Subsequences are selections of terms in order, and the Bolzano–Weierstrass theorem guarantees every bounded real sequence has a convergent subsequence. These ideas, formalized by Cauchy and Weierstrass in the 19th century, are the bedrock of analysis.

💡 The Core Insight

Convergence is an external condition (closeness to a fixed target $L$ ), while the Cauchy condition is an internal one (mutual closeness of terms). The theorem that the two are equivalent — completeness — is what makes $\mathbb{R}$ a usable arena for analysis. Subsequences let us extract order out of chaos: even when a sequence wanders, boundedness forces some thread of it to settle down.

DefinitionSequence

A sequence in a set $X$ is a function $a:\mathbb{N}\to X$ , denoted $(a_n)_{n\ge 1}$ where $a_n = a(n)$ . We mostly take $X=\mathbb{R}$ or $X=\mathbb{C}$ . The range $\{a_n : n\in\mathbb{N}\}$ is a set (may be finite), while the sequence itself is an ordered, possibly-repeating list. A sequence is bounded if there exists $M>0$ with $|a_n|\le M$ for all $n$ .

DefinitionConvergence (

\varepsilon

–

N

)

A sequence $(a_n)$ in $\mathbb{R}$ converges to $L\in\mathbb{R}$ , written $a_n\to L$ or $\lim_{n\to\infty}a_n = L$ , if:

1

\quad \forall\,\varepsilon>0\ \ \exists\,N\in\mathbb{N}\ \ \forall\,n\ge N:\ |a_n - L| < \varepsilon.

Derivation hint: read left to right — given a tolerance $\varepsilon$ , you can name a threshold index $N$ past which all terms lie within $\varepsilon$ of $L$ . Here $\varepsilon$ is an arbitrary positive tolerance, $N$ the threshold index (depends on $\varepsilon$ ), and $|a_n-L|$ the distance from term to limit. A sequence that does not converge diverges.

DefinitionCauchy Sequence

A sequence $(a_n)$ is Cauchy if:

2

\quad \forall\,\varepsilon>0\ \ \exists\,N\in\mathbb{N}\ \ \forall\,m,n\ge N:\ |a_m - a_n| < \varepsilon.

Derivation hint: same structure as [1], but the distance is between two terms of the sequence — no external limit $L$ appears.

DefinitionSubsequence

Given $(a_n)$ and a strictly increasing function $k\mapsto n_k$ (so $n_1<n_2<n_3<\cdots$ , all in $\mathbb{N}$ ), the sequence

3

\quad (a_{n_k})_{k\ge 1} = a_{n_1}, a_{n_2}, a_{n_3}, \dots

Derivation hint: $n_k\ge k$ always, by induction on the strict increase of the index map. is a subsequence of $(a_n)$ . A point $L$ is a subsequential limit (or cluster point) if some subsequence converges to $L$ .

TheoremFundamental Theorems on Sequences

Let $(a_n)$ be a real sequence. 1. (Uniqueness) If $a_n\to L$ and $a_n\to L'$ , then $L=L'$ . 2. (Convergent $\Rightarrow$ bounded) If $(a_n)$ converges, it is bounded. 3. (Convergent $\Rightarrow$ Cauchy) Every convergent sequence is Cauchy. 4. (Cauchy $\Rightarrow$ bounded) Every Cauchy sequence is bounded. 5. (Completeness of $\mathbb{R}$ ) Every Cauchy sequence in $\mathbb{R}$ converges. 6. (Monotone Convergence) A monotone bounded sequence converges (to its sup if increasing, inf if decreasing). 7. (Bolzano–Weierstrass) Every bounded real sequence has a convergent subsequence. 8. (Subsequence inheritance) If $a_n\to L$ , then every subsequence $a_{n_k}\to L$ . 9. (Cluster characterization) $a_n\to L$ iff $(a_n)$ is bounded and $L$ is its only subsequential limit.

ProofConvergent

\Rightarrow

Cauchy (Theorem 3)

1. Suppose $a_n\to L$ . Fix $\varepsilon>0$ . 2. Apply [1] with tolerance $\varepsilon/2$ : there is $N$ such that $n\ge N\Rightarrow |a_n - L| < \varepsilon/2$ . 3. Take any $m,n\ge N$ . By the triangle inequality:

4

\quad |a_m - a_n| = |(a_m - L) - (a_n - L)| \le |a_m - L| + |a_n - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon.

Derivation hint: insert $-L+L$ and split via $|x+y|\le|x|+|y|$ ; each piece is below $\varepsilon/2$ . 4. Thus [2] holds, so $(a_n)$ is Cauchy. $\blacksquare$

∎

ProofBolzano–Weierstrass (Theorem 7) via bisection

1. Let $(a_n)$ satisfy $|a_n|\le M$ , so all terms lie in $I_0 = [-M, M]$ . 2. Bisect $I_0$ into two halves; at least one half contains $a_n$ for infinitely many $n$ . Call it $I_1$ ; pick $n_1$ with $a_{n_1}\in I_1$ . 3. Inductively bisect $I_k$ , choosing the half $I_{k+1}$ containing infinitely many terms, and pick $n_{k+1}>n_k$ with $a_{n_{k+1}}\in I_{k+1}$ . This is possible because $I_{k+1}$ holds infinitely many terms. 4. The interval lengths satisfy

5

\quad |I_k| = \frac{2M}{2^k} = \frac{M}{2^{k-1}} \xrightarrow[k\to\infty]{} 0.

Derivation hint: each bisection halves the length; start from length $2M$ . 5. The nested closed intervals $I_0\supseteq I_1\supseteq\cdots$ with lengths $\to 0$ have, by the Nested Interval Property, a unique common point $L=\bigcap_k I_k$ . 6. Since $a_{n_k}, L\in I_k$ , we get $|a_{n_k}-L|\le |I_k|\to 0$ , so $a_{n_k}\to L$ . $\blacksquare$

∎

ProofCauchy

\Rightarrow

Convergent in

\mathbb{R}

(Theorem 5)

1. Let $(a_n)$ be Cauchy. By Theorem 4 it is bounded. 2. By Bolzano–Weierstrass it has a subsequence $a_{n_k}\to L$ for some $L$ . 3. Fix $\varepsilon>0$ . Choose $N_1$ so that $m,n\ge N_1\Rightarrow |a_m-a_n|<\varepsilon/2$ (Cauchy). Choose $k$ large so that $n_k\ge N_1$ and $|a_{n_k}-L|<\varepsilon/2$ (subsequence limit). 4. For any $n\ge N_1$ :

6

\quad |a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon.

Derivation hint: triangle inequality bridging the term, a deep subsequence term, and the limit. 5. Hence $a_n\to L$ . The boundedness in step 1 plus B–W requires the completeness of $\mathbb{R}$ (Nested Interval Property), which fails in $\mathbb{Q}$ . $\blacksquare$

∎

ProofCauchy

\Rightarrow

bounded (Theorem 4)

1. Apply [2] with $\varepsilon=1$ : there is $N$ with $|a_m-a_n|<1$ for $m,n\ge N$ . 2. In particular $|a_n - a_N| < 1$ , so $|a_n| < |a_N| + 1$ for all $n\ge N$ . 3. Let $M = \max\{|a_1|,\dots,|a_{N-1}|,\ |a_N|+1\}$ . Then $|a_n|\le M$ for all $n$ . $\blacksquare$

∎

mathematician · Augustin-Louis Cauchy | 1789–1857

Cauchy gave analysis its modern rigor in his Cours d'analyse (1821), introducing the convergence criterion now bearing his name and insisting on $\varepsilon$ -style bounds where his predecessors relied on intuition about "infinitely small" quantities. Though his own proof that Cauchy sequences converge implicitly assumed completeness of $\mathbb{R}$ (a property not yet axiomatized), his formulation isolated exactly the right internal condition. Weierstrass and later Cantor and Dedekind supplied the missing foundation by constructing $\mathbb{R}$ so that Cauchy sequences would necessarily converge.

🧠 Intuition

Think of convergence as a homing missile: $L$ is the target, and for any shrinking circle of radius $\varepsilon$ around it, the sequence eventually enters and never leaves. The Cauchy condition removes the target: instead, you watch the terms huddle together — past some point they all crowd into a region of diameter $\varepsilon$ . In $\mathbb{R}$ , huddling guarantees a home: there's always a real number for the huddle to converge to. In $\mathbb{Q}$ , the huddle can form around a "hole" (an irrational), and no rational target exists — the sequence is Cauchy but stranded. A subsequence is choosing which terms to watch, in order, skipping freely; Bolzano–Weierstrass says a bounded sequence may bounce around, but you can always pick out a calm thread that settles.

Figure

Convergence funnel: terms enter the

\varepsilon

-band past threshold

N

ExampleProving

a_n = \tfrac{2n+1}{n+3}\to 2

by

\varepsilon

–

N

Given: $a_n = \dfrac{2n+1}{n+3}$ , candidate limit $L=2$ , target tolerance demonstration for general $\varepsilon>0$ .

Step 1 — compute the distance.

7

\quad |a_n - 2| = \left|\frac{2n+1}{n+3} - 2\right| = \left|\frac{2n+1 - 2n - 6}{n+3}\right| = \frac{5}{n+3}.

Derivation hint: combine over a common denominator; numerator simplifies to $-5$ .

Step 2 — solve $\dfrac{5}{n+3} < \varepsilon$ . This holds iff $n+3 > 5/\varepsilon$ , i.e. $n > \dfrac{5}{\varepsilon} - 3$ .

Step 3 — choose $N$ . Take $N = \left\lceil \dfrac{5}{\varepsilon} \right\rceil$ (a safe choice, since $N > 5/\varepsilon - 3$ ). Then for all $n\ge N$ , $|a_n-2| = \frac{5}{n+3} < \varepsilon$ .

Numerical check: for $\varepsilon = 0.01$ , $N = \lceil 500\rceil = 500$ . At $n=500$ : $a_{500} = \frac{1001}{503} \approx 1.98807$ , and $|a_{500}-2|\approx 0.01193$ ... slightly over, so we verify $n=503$ : $a_{503}=\frac{1007}{506}\approx 1.99012$ , $|{\cdot}|\approx 0.00988<0.01$ . ✓ (the $-3$ slack confirms the bound is comfortably satisfied past $N$ ).

\boxed{\ \lim_{n\to\infty}\frac{2n+1}{n+3} = 2,\quad N(\varepsilon)=\left\lceil \tfrac{5}{\varepsilon}\right\rceil\ }

ExampleA Cauchy sequence in

\mathbb{Q}

with no rational limit

Given: Define $(a_n)$ by truncating the decimal expansion of $\sqrt 2 = 1.41421356\dots$ : $a_1=1.4,\ a_2=1.41,\ a_3=1.414,\dots$ , so each $a_n$ has $n$ decimal places.

Step 1 — show Cauchy. For $m>n$ , both share the first $n$ digits, so they differ only beyond the $n$ -th decimal place:

8

\quad |a_m - a_n| \le 10^{-n}.

Derivation hint: truncations agree to $n$ places; the maximal disagreement is in the $(n{+}1)$ th place onward, bounded by a geometric tail $\sum_{k>n}9\cdot10^{-k}=10^{-n}$ .

Step 2 — verify [2]. Given $\varepsilon>0$ , choose $N$ with $10^{-N}<\varepsilon$ , i.e. $N > \log_{10}(1/\varepsilon)$ . For $m,n\ge N$ , $|a_m-a_n|\le 10^{-\min(m,n)}\le 10^{-N}<\varepsilon$ . So $(a_n)$ is Cauchy in $\mathbb{Q}$ .

Step 3 — no rational limit. If $a_n\to q\in\mathbb{Q}$ , then since $a_n^2\to 2$ (and $a_n^2$ rational), $q^2=2$ — impossible for rational $q$ (classical irrationality of $\sqrt 2$ ).

Numerical check: $a_5 = 1.41421$ , $a_8 = 1.41421356$ , $|a_8-a_5| = 0.00000356 < 10^{-5}$ . ✓ The terms huddle, but the home $\sqrt 2$ lives outside $\mathbb{Q}$ .

\boxed{\ (a_n)\text{ is Cauchy in }\mathbb{Q}\text{ but diverges in }\mathbb{Q};\ \text{converges to }\sqrt2\text{ in }\mathbb{R}\ }

⚠️ Common mistake

"Cauchy always means convergent." False in a general metric space. Cauchy $\Rightarrow$ convergent only in complete spaces. The implication is the definition of completeness, not a free theorem. In $\mathbb{Q}$ , $(0,1]$ with usual metric, or the space of continuous functions under the $L^1$ norm, Cauchy sequences can fail to converge.

" $|a_{n+1}-a_n|\to 0$ implies Cauchy." Decisively false. Consecutive gaps shrinking does not control cumulative drift over far-apart indices.

CounterexampleConsecutive differences vanish, yet the sequence diverges

Let $a_n = \sum_{k=1}^{n}\frac{1}{k}$ (harmonic partial sums). Then

9

\quad |a_{n+1}-a_n| = \frac{1}{n+1}\to 0,\qquad\text{but}\qquad a_{2n}-a_n = \sum_{k=n+1}^{2n}\frac1k \ge n\cdot\frac{1}{2n} = \frac12.

Derivation hint: each of the $n$ terms in the block exceeds $1/(2n)$ ; summing gives a fixed lower bound $1/2$ , blocking the Cauchy condition for $\varepsilon=1/2$ . Since $|a_{2n}-a_n|\not\to 0$ , $(a_n)$ is not Cauchy, hence diverges (in fact $a_n\to\infty$ ). The vanishing of consecutive gaps is far too weak.

CounterexampleA bounded divergent sequence — with two subsequential limits

$a_n = (-1)^n$ is bounded by $1$ but does not converge: the even-indexed subsequence $a_{2k}=1\to 1$ while the odd-indexed $a_{2k+1}=-1\to -1$ . Two distinct subsequential limits violate Theorem 8 if the sequence converged, so it cannot. This illustrates that Bolzano–Weierstrass guarantees a convergent subsequence (here two), not convergence of the whole sequence.

connection

Sequence convergence is the engine beneath nearly all of analysis. Series and Convergence Tests defines a series as the sequence of partial sums, so every convergence test is secretly a sequence statement. Limits of Functions can be characterized sequentially (Heine's criterion: $f(x)\to L$ as $x\to a$ iff $f(x_n)\to L$ for every sequence $x_n\to a$ ). Continuity then becomes "sequence-preserving." The Cauchy criterion generalizes to Metric Spaces, where completeness distinguishes well-behaved spaces, and to Banach Spaces (complete normed vector spaces) underlying functional analysis. Bolzano–Weierstrass is the prototype of sequential compactness, linking to Compactness and the Heine–Borel theorem.

🔬 Where this appears

⏳ History

- c. 450 BCE — Zeno's paradoxes implicitly probe infinite sequences and sums, foreshadowing limit puzzles.

- c. 1665 — Newton and Leibniz use convergent series and "ultimate ratios" in calculus, without rigorous limits.

- 1734 — Berkeley's The Analyst attacks the logical vagueness of infinitesimals ("ghosts of departed quantities"), exposing the need for rigor.

- 1817 — Bolzano proves (in obscurity) an intermediate-value-style result and the essence of what becomes the Bolzano–Weierstrass theorem.

- 1821 — Cauchy's Cours d'analyse formalizes convergence and the Cauchy criterion, anchoring analysis in inequalities.

- 1872 — Weierstrass, Cantor, Dedekind, and Heine independently rigorize the real numbers; Cantor builds

\mathbb{R}

from Cauchy sequences; Dedekind from cuts.

- 1872 — Weierstrass popularizes the

\varepsilon

–

\delta

/

\varepsilon

–

N

formalism in lectures, the form taught today.

- 1906 — Fréchet introduces abstract metric spaces, generalizing Cauchy sequences and completeness beyond

\mathbb{R}

.

- 1922 — Banach's thesis establishes complete normed spaces and the contraction mapping principle, embedding sequence convergence at the heart of functional analysis.

open-problem

Completeness is space-dependent, not absolute. Whether a Cauchy sequence converges depends entirely on the ambient space; $(\mathbb{Q},|\cdot|)$ is incomplete, but $\mathbb{Q}$ can be re-metrized $p$ -adically (the $p$ -adic absolute value) so that entirely different sequences become Cauchy, yielding the field $\mathbb{Q}_p$ — a separate, equally valid completion. Ostrowski's theorem classifies all completions of $\mathbb{Q}$ : up to equivalence, exactly the real and the $p$ -adic ones. Open and active areas include the geometry of completions in non-Archimedean analysis, convergence in non-metrizable topological spaces (where sequences must be replaced by nets/filters because they no longer detect closure), and the constructive/computable content of completeness — Bishop-style constructive analysis questions which existence proofs (like Bolzano–Weierstrass) yield algorithms, since the classical bisection proof is not fully constructive without extra choice principles.

Sequences: Convergence, Cauchy Sequences, Subsequences

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also