The Cauchy Strategy

When the Limit Is Unknown

The Infimum Strategy identifies a candidate by proving it is a sharp lower bound. Some convergence problems offer no comparably natural candidate: a sequence is defined by a recurrence, an iterative procedure, or partial sums, and its limit is not apparent. In these situations, it can be more effective to compare late terms with one another than to compare them with an unknown limit.

That shift is the Cauchy Strategy. Rather than first proposing a number \(L\) and proving \(x_n\) approaches it, show that all sufficiently late terms are close to each other. Completeness then supplies a limit. For real sequences, the Cauchy Criterion says that a sequence converges if and only if it is Cauchy. The forward direction follows from the triangle inequality; the reverse direction uses the completeness of \(\mathbb{R}\). Thus a Cauchy proof can establish convergence even when it does not identify the limit.

Definition (Cauchy Sequence): A sequence \((x_n)\) in a metric space \((X,d)\) is Cauchy if, for every \(\varepsilon>0\), there exists an integer \(N\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\geq N\).

The quantifiers are important. One index \(N\), chosen for the given \(\varepsilon\), must work for every pair of indices \(m,n\) in the tail. The Cauchy condition asks for control of the entire tail, not merely for consecutive terms to be close.

For real sequences, the Cauchy Criterion gives the main bridge from this definition to convergence. We will use completeness of \(\mathbb{R}\) for that bridge, and develop two practical tools: a tail estimate from summable successive differences, and a convergence proof for contraction iterations.

From Successive Differences to a Tail Estimate

When a sequence is constructed step by step, its successive differences may be easier to estimate than the distance between two arbitrary terms. The triangle inequality connects these two kinds of information: the distance from \(x_n\) to \(x_m\) is at most the sum of the intervening successive distances. If those bounds have a finite total sum, their tails become arbitrarily small.

Theorem (Summable Successive Differences Give a Cauchy Sequence): Let \((X,d)\) be a metric space, and let \((x_n)\) be a sequence in \(X\). Suppose there are nonnegative real numbers \(a_k\) such that \(\sum_{k=1}^{\infty}a_k\) converges and \(d(x_{k+1},x_k)\leq a_k\) for every \(k\geq1\). Then \((x_n)\) is Cauchy.

Proof. Let \(\varepsilon>0\). Since the series of nonnegative terms \(\sum_{k=1}^{\infty}a_k\) converges, its tails tend to zero. Choose \(N\) such that \(\sum_{k=N}^{\infty}a_k<\varepsilon\). Take any \(m,n\geq N\). If \(m>n\), repeated use of the triangle inequality gives

d(x_m,x_n)\leq\sum_{k=n}^{m-1}d(x_{k+1},x_k) \leq\sum_{k=n}^{m-1}a_k \leq\sum_{k=N}^{\infty}a_k <\varepsilon.

If \(n>m\), the same estimate applies with the indices reversed, since \(d(x_m,x_n)=d(x_n,x_m)\). If \(m=n\), then \(d(x_m,x_n)=0<\varepsilon\). Thus every pair \(m,n\geq N\) satisfies \(d(x_m,x_n)<\varepsilon\), which is the Cauchy condition. \(\square\)

Worked Example: A Sequence With Alternating Signs

Consider \(x_n=(-1)^n/n\) for positive integers \(n\). To prove convergence without initially needing a candidate, let \(\varepsilon>0\) and choose \(N>2/\varepsilon\). For \(m,n\geq N\), the triangle inequality gives

|x_m-x_n|\leq |x_m|+|x_n| =\frac{1}{m}+\frac{1}{n} \leq\frac{2}{N}<\varepsilon.

So \((x_n)\) is Cauchy and therefore converges in \(\mathbb{R}\). In this example, the estimate also shows directly that \(x_n\to0\): for \(n\geq N\), \(|x_n|=1/n\leq1/N<\varepsilon\). The Cauchy argument is useful even when such a recognizable candidate is unavailable.

Worked Example: Convergence of a Sequence of Partial Sums

Define \(s_n=\sum_{k=1}^{n}1/(k(k+1))\). To compare any two terms, take \(m>n\). The identity \(1/(k(k+1))=1/k-1/(k+1)\) gives the finite tail estimate

|s_m-s_n| =\sum_{k=n+1}^{m}\frac{1}{k(k+1)} =\frac{1}{n+1}-\frac{1}{m+1} <\frac{1}{n+1}.

Given \(\varepsilon>0\), choose \(N\) such that \(1/(N+1)<\varepsilon\). If \(m>n\geq N\), the displayed estimate is less than \(\varepsilon\); if \(n>m\geq N\), interchange the indices; and if \(m=n\), the difference is zero. Hence \((s_n)\) is Cauchy and converges in \(\mathbb{R}\).

Here the same calculation identifies the limit if desired. The finite sum telescopes to \(s_n=1-1/(n+1)\), which tends to \(1\). The point of the Cauchy proof is that it required only a bound on the tail between two partial sums; it did not need the limit in advance.

A Complete-Space Result for Iteration

A particularly useful application arises when a map repeatedly brings points closer together. If each iteration contracts distances by a fixed factor less than one, then the successive steps shrink geometrically. The summable-differences theorem turns that geometric estimate into a Cauchy sequence.

Theorem (Contraction Iteration Converges in a Complete Metric Space): Let \((X,d)\) be a nonempty complete metric space, and let \(T:X\to X\) satisfy \(d(Tx,Ty)\leq q\,d(x,y)\) for all \(x,y\in X\), where \(0\leq q<1\). For any \(x_0\in X\), define \(x_{n+1}=T(x_n)\). Then \((x_n)\) converges to a point \(x^*\in X\) satisfying \(T(x^*)=x^*\). This fixed point is unique.

Proof. If \(q=0\), the contraction inequality gives \(d(Tx,Ty)=0\) for all \(x,y\), so \(T\) is constant. Then \(x_n=x_1\) for every \(n\geq1\), and \(x_1=T(x_1)\), which proves convergence and gives a fixed point. Any fixed point must equal the value of the constant map, so it is unique.

Now suppose \(0<q<1\). Put \(D=d(x_1,x_0)\). By repeated application of the contraction inequality,

d(x_{k+1},x_k)\leq q^kD\qquad(k\geq0).

Indeed, the case \(k=0\) is equality. If the estimate holds at \(k\), then

d(x_{k+2},x_{k+1}) =d(Tx_{k+1},Tx_k) \leq q\,d(x_{k+1},x_k) \leq q^{k+1}D.

For integers \(m>n\geq0\), the triangle inequality and the finite geometric sum give

d(x_m,x_n) \leq\sum_{k=n}^{m-1}d(x_{k+1},x_k) \leq D\sum_{k=n}^{m-1}q^k \leq \frac{Dq^n}{1-q}.

Since \(q^n\to0\), this bound proves that \((x_n)\) is Cauchy. Completeness of \(X\) gives a point \(x^*\in X\) with \(x_n\to x^*\). The contraction inequality implies continuity of \(T\): \(d(Tx_n,Tx^*)\leq qd(x_n,x^*)\to0\). But \(Tx_n=x_{n+1}\), and the shifted sequence also converges to \(x^*\). Therefore \(Tx^*=x^*\).

Finally, if \(u\) and \(v\) are fixed points, then \(d(u,v)=d(Tu,Tv)\leq qd(u,v)\). Thus \((1-q)d(u,v)\leq0\). Since \(1-q>0\) and distances are nonnegative, \(d(u,v)=0\), so \(u=v\). The fixed point is unique. \(\square\)

Worked Example: Iterating a Contraction on the Real Line

Let \(T:\mathbb{R}\to\mathbb{R}\) be \(T(x)=(x+4)/3\), and start at \(x_0=-1\). For real \(x,y\),

|T(x)-T(y)|=\left|\frac{x-y}{3}\right|=\frac{1}{3}|x-y|.

Thus \(T\) is a contraction with \(q=1/3\), and \(\mathbb{R}\) is complete. The theorem ensures that the iterates converge to a fixed point. Solving \(T(x)=x\) gives \((x+4)/3=x\), hence \(x=2\).

The estimates can also be checked explicitly. Since \(x_1=T(-1)=1\), induction gives \(x_n=2-3/3^n\) for \(n\geq0\): it holds at \(n=0\), and if \(x_n=2-3/3^n\), then \(T(x_n)=(2-3/3^n+4)/3=2-3/3^{n+1}\). Therefore \(x_n\to2\), consistent with the Cauchy argument and the fixed-point equation.

Use Tail Bounds Carefully

A useful Cauchy proof makes the dependence on the chosen error visible. First decide which two late terms must be compared. Then bound their distance by a quantity controlled by the smaller index, or by a tail of a convergent series. Finally, choose one threshold \(N\) that makes this bound smaller than \(\varepsilon\). For a sequence in \(\mathbb{R}\), once the Cauchy condition is established, invoke the Cauchy Criterion to conclude convergence.

A common mistake is to prove only that \(d(x_{n+1},x_n)\to0\). This does not imply that the sequence is Cauchy: small successive steps can accumulate over many indices. The summable-differences theorem requires the stronger information that the step bounds have a convergent sum. Another frequent error is to choose \(N\) depending on \(m\) or \(n\); the Cauchy definition requires a single \(N\) that works for every pair in the tail.

Write the two-index target.
Fix arbitrary \(m,n\geq N\) and plan to bound the distance between those terms.

Build a tail estimate.
Use the triangle inequality to sum successive differences, or estimate the tail directly.

Choose the threshold.
Make the bound smaller than the given \(\varepsilon\) with an \(N\) independent of the two indices.

Invoke completeness where needed.
A Cauchy sequence converges in a complete space; use continuity afterward if the limit must satisfy an equation.

The Cauchy Strategy is most powerful when the limit is difficult to guess but the tail is controllable. Its central discipline is to prove a uniform estimate for every pair of sufficiently late terms. That estimate creates the Cauchy property; completeness then converts the internal consistency of the tail into an actual limit.

Check Your Understanding

Use the definitions and arguments in this tutorial to answer these questions.

What is the difference between saying successive differences tend to zero and proving that a sequence is Cauchy?
Why does a convergent sum of nonnegative step bounds make the corresponding sequence Cauchy?
Where is completeness used in the contraction iteration theorem?
Why does continuity of the contraction imply that the limit of the iterates is a fixed point?
How does the proof establish uniqueness of the fixed point?

What You'll Learn