Cauchy sequences; completeness of R

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2020, 2024, 2025)
• Priority tier: T3
• Marks (count): 10 (1), 15 (2)
• Average solve time: ~10 min
• Difficulty mix: easy 1, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Cauchy sequences are the backbone of completeness and appear consistently at 10–15 marks. The three question types are: (1) prove a contractive-type sequence is Cauchy by geometric telescoping (2020); (2) apply Cauchy’s general principle to establish convergence of a concrete series (2024); (3) define Cauchy sequences and prove every convergent sequence is Cauchy, then discuss completeness (2025). All three have tight, reproducible proof templates. The 2025 question is the most theoretical — mastering it gives the clearest understanding of why completeness distinguishes $\mathbb{R}$ from $\mathbb{Q}$.

Minimum Theory

Definition. A sequence $(a_n)$ in $\mathbb{R}$ is Cauchy if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$. The definition involves the terms themselves, with no reference to a limit value.

Cauchy criterion (completeness of $\mathbb{R}$). In $\mathbb{R}$, a sequence is convergent if and only if it is Cauchy. The forward direction (convergent $\Rightarrow$ Cauchy) follows from the $\varepsilon/2$-triangle argument. The backward direction (Cauchy $\Rightarrow$ convergent) is the completeness of $\mathbb{R}$; $\mathbb{Q}$ is not complete (decimal truncations of $\sqrt{2}$ are Cauchy in $\mathbb{Q}$ but have no rational limit).

Geometric telescoping. If $|a_{n+1} - a_n| \le \alpha^{n-1} D$ with $0 < \alpha < 1$ and $D = |a_2 - a_1|$, then for $m > n$:

$|a_m - a_n| \le \sum_{k=n}^{m-1}|a_{k+1}-a_k| \le D\alpha^{n-1}\sum_{j=0}^{\infty}\alpha^j = \frac{D\alpha^{n-1}}{1-\alpha} \to 0.$

This bound is independent of $m$, so choosing $N$ with $\alpha^{N-1} < \varepsilon(1-\alpha)/D$ completes the Cauchy argument.

Question Archetypes

cauchy-contraction (1 question; 2020)

Recognition Cues — The hypothesis gives a contractive inequality $|a_{n+1} - a_n| \le \alpha |a_n - a_{n-1}|$ with $0 < \alpha < 1$ for all $n \ge 2$. The question asks to prove $(a_n)$ is Cauchy. No explicit formula for $a_n$ is given — the proof is purely from the contraction estimate.

Solution Template

1. Set $d_n = |a_{n+1} - a_n|$; iterate the contraction to get $d_n \le \alpha^{n-1} d_1$.
2. For $m > n$, telescope: $|a_m - a_n| \le \sum_{k=n}^{m-1} d_k \le \frac{D \alpha^{n-1}}{1-\alpha}$ (bound independent of $m$).
3. Note $\alpha^{n-1} \to 0$ so the bound $\to 0$.
4. Given $\varepsilon > 0$, choose $N$ so $\frac{D\alpha^{N-1}}{1-\alpha} < \varepsilon$; conclude Cauchy.

Worked Example

2020 Paper 2, 2020-P2-Q1c (10 marks)

Prove that the sequence $(a_n)$ satisfying $|a_{n+1} - a_n| \le \alpha |a_n - a_{n-1}|$, $0 < \alpha < 1$, for all $n \ge 2$, is a Cauchy sequence.

Step 1 — Geometric bound on consecutive differences.

Let $d_n = |a_{n+1} - a_n|$ for $n \ge 1$. By hypothesis $d_n \le \alpha d_{n-1}$ for $n \ge 2$. Iterating:

$d_n \le \alpha^{n-1} d_1 = \alpha^{n-1} D, \quad D = |a_2 - a_1|.$

If $D = 0$ the sequence is constant from $a_1$, hence trivially Cauchy. Assume $D > 0$.

Step 2 — Triangle inequality bound for $|a_m - a_n|$.

For $m > n \ge 1$, telescoping and the triangle inequality give:

$|a_m - a_n| \le \sum_{k=n}^{m-1}|a_{k+1}-a_k| \le \sum_{k=n}^{m-1}\alpha^{k-1}D = D\alpha^{n-1}\sum_{j=0}^{m-n-1}\alpha^j < \frac{D\alpha^{n-1}}{1-\alpha}.$

The bound $\dfrac{D\alpha^{n-1}}{1-\alpha}$ is independent of $m$.

Step 3 — Cauchy criterion.

Since $0 < \alpha < 1$, $\alpha^{n-1} \to 0$, so $\dfrac{D\alpha^{n-1}}{1-\alpha} \to 0$.

Let $\varepsilon > 0$. Choose $N$ so large that $\dfrac{D\alpha^{N-1}}{1-\alpha} < \varepsilon$. Then for all $m > n \ge N$:

$|a_m - a_n| \le \frac{D\alpha^{n-1}}{1-\alpha} \le \frac{D\alpha^{N-1}}{1-\alpha} < \varepsilon.$

$\boxed{|a_m - a_n| \le \frac{D\,\alpha^{n-1}}{1-\alpha} \to 0 \implies (a_n) \text{ is a Cauchy sequence.}}$

Common Traps

• Sum the geometric series to $1/(1-\alpha)$ — a constant independent of $m$. Leaving an $m$-dependent partial sum means the bound cannot be used to pick a single $N$.
• The condition $\alpha < 1$ is used in two places: summing the geometric series, and sending $\alpha^{n-1} \to 0$. With $\alpha = 1$ neither works (e.g., $a_n = n$ satisfies $d_n = 1 = d_{n-1}$ but is not Cauchy).
• Treat $D = 0$ (sequence is immediately constant) as a trivial base case before the main argument.

cauchy-criterion (1 question; 2024)

Recognition Cues — The sequence $f_n$ is a partial sum of a series with explicitly given terms. The question says “using Cauchy’s general principle, examine the convergence.” You must bound $|f_m - f_n|$ for $m > n$ and show it can be made $< \varepsilon$ independent of $m$.

Solution Template

1. State Cauchy’s general principle explicitly.
2. For $m > n$, write $|f_m - f_n| = \sum_{k=n+1}^{m} |a_k|$.
3. Bound $|a_k|$ by a term of a known convergent series (here: $1/k! \le 1/2^{k-1}$, geometric).
4. Sum the tail to a bound depending only on $n$ (not $m$); show this $\to 0$.
5. Choose $N$ from the tail bound; conclude convergent.

Worked Example

2024 Paper 2, 2024-P2-Q2a (15 marks)

Using Cauchy’s general principle of convergence, examine the convergence of $\langle f_n \rangle$, where $f_n = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \cdots + \dfrac{1}{n!}$.

Cauchy’s general principle. $\langle f_n \rangle$ converges $\iff$ for every $\varepsilon > 0$ there exists $N$ such that $|f_m - f_n| < \varepsilon$ for all $m, n \ge N$.

Step 1 — Bound the tail.

For $m > n$:

$|f_m - f_n| = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots + \frac{1}{m!}.$

Factorial bound: $k! \ge 2^{k-1}$ for all $k \ge 1$ (induction: $1! = 1 = 2^0$; $(k+1)! = (k+1)k! \ge 2 \cdot 2^{k-1} = 2^k$). Hence $1/k! \le 1/2^{k-1}$.

$|f_m - f_n| \le \frac{1}{2^n} + \frac{1}{2^{n+1}} + \cdots + \frac{1}{2^{m-1}} < \sum_{k=n}^{\infty}\frac{1}{2^k} = \frac{1}{2^{n-1}}.$

Step 2 — Verify Cauchy criterion.

The bound $1/2^{n-1}$ is independent of $m$ and $\to 0$ as $n \to \infty$. Given $\varepsilon > 0$, choose $N$ with $1/2^{N-1} < \varepsilon$. Then for all $m > n \ge N$:

$|f_m - f_n| < \frac{1}{2^{n-1}} \le \frac{1}{2^{N-1}} < \varepsilon.$

By Cauchy’s general principle:

$\boxed{\langle f_n \rangle \text{ converges.} \quad (\text{In fact, the limit is } e = \sum_{k=0}^{\infty} 1/k!.)}$

Common Traps

• The bound must hold for all $m \ge n \ge N$ (including $m \gg n$); confirm it is independent of $m$.
• The bound $1/k! \le 1/2^{k-1}$ is the right one; the weaker $1/k! \le 1/k$ does not yield a summable tail.
• Do not prove the series converges by citing $e$ — the question specifically asks for the Cauchy characterisation.

cauchy-convergent-proof (1 question; 2025)

Recognition Cues — Asks you to (a) define a Cauchy sequence, (b) prove every convergent sequence is Cauchy, and (c) discuss the importance of the Cauchy condition. Three clearly labelled parts are expected.

Solution Template

1. Definition: state the $\varepsilon$-$N$ definition of a Cauchy sequence explicitly.
2. Theorem + Proof: convergent $\Rightarrow$ Cauchy via the $\varepsilon/2$ triangle argument.
3. Importance: (i) intrinsic criterion — no need to know the limit; (ii) completeness of $\mathbb{R}$ — Cauchy $\iff$ convergent; (iii) $\mathbb{Q}$ is not complete (example); (iv) foundational role in analysis.

Worked Example

2025 Paper 2, 2025-P2-Q2a (15 marks)

Define Cauchy sequence and prove that every convergent sequence of real numbers is a Cauchy sequence. What is the importance of Cauchy condition?

Definition. A sequence $(a_n)_{n \ge 1}$ is a Cauchy sequence if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

$|a_m - a_n| < \varepsilon \quad \text{for all } m, n \ge N.$

Intuitively: the terms become arbitrarily close to one another, without reference to any limit.

Theorem. Every convergent sequence of real numbers is a Cauchy sequence.

Proof. Suppose $(a_n) \to L \in \mathbb{R}$: for every $\varepsilon > 0$ there exists $N$ with $|a_n - L| < \varepsilon/2$ for all $n \ge N$. For any $m, n \ge N$, the triangle inequality gives:

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

Hence $(a_n)$ is Cauchy. $\blacksquare$

Importance of the Cauchy condition.

1. Intrinsic convergence test. Cauchy’s condition uses only the terms of the sequence — no prior knowledge of the limit. This allows us to establish convergence even when the limit is unknown.

2. Completeness of $\mathbb{R}$. The converse (every Cauchy sequence in $\mathbb{R}$ converges) is the completeness of $\mathbb{R}$. Together:

$\text{a real sequence converges} \iff \text{it is Cauchy.}$

3. $\mathbb{Q}$ is not complete. The decimal truncations of $\sqrt{2}$ form a Cauchy sequence in $\mathbb{Q}$ but converge to an irrational; completeness is a special feature of $\mathbb{R}$.

4. Foundational role. Completeness underlies convergence of series (Cauchy series test), uniform convergence, Cantor’s construction of $\mathbb{R}$ from $\mathbb{Q}$, the Banach fixed-point theorem, and the general theory of Banach spaces.

$\boxed{\text{Real sequence converges} \iff \text{Cauchy} \quad (\text{completeness of } \mathbb{R}).}$

Common Traps

• Use $\varepsilon/2$ (not $\varepsilon$) in the triangle inequality step; otherwise the bound gives $2\varepsilon$ and a separate $\varepsilon \mapsto \varepsilon/2$ substitution is needed.
• The importance section must include completeness and a $\mathbb{Q}$-counterexample for full marks; omitting it loses the third of the answer.

Marks-Aware Writing

10-mark answer (cauchy-contraction): Four steps — geometric iteration, telescoping bound, send to zero, $\varepsilon$-$N$ conclusion. Each step needs one or two lines of algebra. Handle $D=0$ in one sentence.

15-mark answer (cauchy-criterion): State Cauchy’s principle first (2 lines), establish $k! \ge 2^{k-1}$ by induction (4 lines), bound $|f_m - f_n| < 1/2^{n-1}$ (3 lines), choose $N$ and close (3 lines).

15-mark answer (cauchy-convergent-proof): Three sections of roughly equal length — definition (3 lines), theorem + proof (5 lines), importance (6–8 lines covering at least 3 points). The importance section carries roughly a third of the marks.

Practice Set

• 2020-P2-Q1c (10 m) — — Hint: iterate $d_n \le \alpha d_{n-1}$ to $d_n \le \alpha^{n-1}D$; the final bound on $|a_m - a_n|$ must be free of $m$.
• 2024-P2-Q2a (15 m) — — Hint: use $k! \ge 2^{k-1}$ to reduce to a geometric tail; state Cauchy’s principle first.
• 2025-P2-Q2a (15 m) — — Hint: three-part structure — definition, $\varepsilon/2$ proof, importance with at least three enumerated points.