Prove the Cauchy Criterion

Convergence Without Knowing the Limit

The Monotone Convergence Theorem proves that a bounded monotone sequence converges, but many sequences are not monotone. A different question can be easier to answer: do the terms eventually become close to one another? If they do, there may be no obvious candidate for a limit, yet the sequence still has a strong form of internal stability.

The Cauchy criterion makes this idea precise. Its two directions have different roles. Any convergent sequence must have terms close to one another far enough out. Conversely, for real sequences, terms that become arbitrarily close to one another must approach a real limit. The second direction depends on the completeness of \(\mathbb R\), and its proof will use the Monotone Convergence Theorem from the previous tutorial.

Definition: A real sequence \((a_n)\) is Cauchy if for every \(\varepsilon>0\), there is a positive integer \(N\) such that \(|a_m-a_n|<\varepsilon\) whenever \(m,n\geq N\).

Both indices in this definition must be at least \(N\). The condition is not merely that consecutive terms become close: it requires every pair of terms in the tail to be close. The criterion can be written without referring to any proposed limit, which is its main practical advantage.

Every Convergent Sequence Is Cauchy

Theorem (Convergence Implies the Cauchy Condition): Every convergent real sequence is Cauchy.

Proof. Suppose \(a_n\to L\), and let \(\varepsilon>0\). By the definition of convergence, there is a positive integer \(N\) such that \(|a_k-L|<\varepsilon/2\) whenever \(k\geq N\). If \(m,n\geq N\), the triangle inequality gives

|a_m-a_n| \leq |a_m-L|+|a_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

Thus every pair of terms with indices at least \(N\) differs by less than \(\varepsilon\). This is exactly the Cauchy condition. \(\square\)

The use of \(\varepsilon/2\) leaves room for the two errors to add while staying below \(\varepsilon\). The same idea works in any metric space: if two points of a sequence are each within \(\varepsilon/2\) of the same limit, their distance from one another is less than \(\varepsilon\).

Worked Example: A Reciprocal Sequence Is Cauchy

Consider \(a_n=1/(n+1)\). Let \(\varepsilon>0\). Choose a positive integer \(N\) such that \(N+1>1/\varepsilon\). If \(m,n\geq N\), then \(m+1\geq N+1\) and \(n+1\geq N+1\), so both terms are positive and at most \(1/(N+1)\). Therefore

|a_m-a_n| \leq \max\{a_m,a_n\} \leq \frac{1}{N+1} <\varepsilon.

This verifies the Cauchy condition directly. The argument compares arbitrary terms in the tail and does not need to begin by identifying a limit.

Tail Supremum and Infimum

To prove the converse for real sequences, we will track the entire range of each tail. For a bounded sequence, each tail has a supremum and an infimum. As the tail advances, its supremum cannot increase, and its infimum cannot decrease. These two sequences of bounds are therefore monotone.

A Cauchy sequence is bounded, by the result Every Cauchy Sequence of Real Numbers Is Bounded from the Contradiction Strategy tutorial. Consequently, all its tails are bounded, so their suprema and infima exist by completeness of the real numbers. The Monotone Convergence Theorem then gives limits for these bounds.

Theorem (Cauchy Criterion for Real Sequences): A real sequence converges if and only if it is Cauchy.

Proof. The forward direction is the theorem just proved. For the converse, suppose \((a_n)\) is Cauchy. For each positive integer \(n\), define its tail set and tail bounds by

A_n=\{a_k:k\geq n\},\qquad u_n=\sup A_n,\qquad \ell_n=\inf A_n.

Each \(A_n\) is nonempty. Since the sequence is bounded, \(A_n\) is bounded above and below, so \(u_n\) and \(\ell_n\) are real numbers. The inclusion \(A_{n+1}\subseteq A_n\) implies \(u_{n+1}\leq u_n\) and \(\ell_n\leq\ell_{n+1}\). Thus \((u_n)\) is nonincreasing and bounded below, while \((\ell_n)\) is nondecreasing and bounded above. By the decreasing and increasing cases of the Monotone Convergence Theorem, there are real numbers \(u\) and \(\ell\) such that \(u_n\to u\) and \(\ell_n\to\ell\).

We next show that these two limits are equal. Let \(\eta>0\). Since \((a_n)\) is Cauchy, there is an \(N\) such that \(|a_j-a_k|<\eta\) for all \(j,k\geq N\). In particular, \(a_j<a_k+\eta\) for all such \(j,k\). Fixing \(k\geq N\), every element of \(A_N\) is at most \(a_k+\eta\), so \(u_N\leq a_k+\eta\). This holds for every \(k\geq N\), which means \(u_N-\eta\) is a lower bound for \(A_N\). Hence \(u_N-\eta\leq\ell_N\), or

0\leq u_N-\ell_N\leq\eta.

For every \(n\geq N\), the same argument applies to the smaller tail \(A_n\), so \(0\leq u_n-\ell_n\leq\eta\). Since \(u_n\to u\) and \(\ell_n\to\ell\), order is preserved under limits, giving \(0\leq u-\ell\leq\eta\). This holds for every \(\eta>0\), so \(u=\ell\).

Finally, the definitions of supremum and infimum give \(\ell_n\leq a_n\leq u_n\) for every \(n\). Both bounding sequences converge to the same real number \(u=\ell\). The Squeeze Theorem therefore gives \(a_n\to u\). This proves that every real Cauchy sequence converges and completes the proof. \(\square\)

The tail bounds are useful because they capture all the terms from a given index onward, not just one selected term. The Cauchy condition makes the width between the tail’s supremum and infimum arbitrarily small. Monotonicity supplies limits for the two bounds, and their vanishing separation forces those limits to agree.

Worked Applications

Worked Example: Geometric Partial Sums Are Cauchy

For \(n\geq1\), let \(s_n=\sum_{k=1}^{n}1/4^k\). If \(m>n\), the geometric-sum formula gives

|s_m-s_n| =\sum_{k=n+1}^{m}\frac{1}{4^k} =\frac{1}{3}\left(\frac{1}{4^n}-\frac{1}{4^m}\right) <\frac{1}{3\cdot4^n}.

The formula follows by multiplying the finite sum by \(1-1/4=3/4\), which leaves \(1/4^{n+1}-1/4^{m+1}\), and then dividing by \(3/4\). If \(n>m\), interchanging \(m\) and \(n\) gives the same bound with the smaller index. Given \(\varepsilon>0\), choose \(N\) such that \(1/(3\cdot4^N)<\varepsilon\). For any \(m,n\geq N\), the difference is zero if \(m=n\); otherwise the bound above, using the smaller index, is at most \(1/(3\cdot4^N)<\varepsilon\). Hence \((s_n)\) is Cauchy.

Worked Example: Alternating Terms Do Not Form a Cauchy Sequence

Let \(a_n=(-1)^n\). To show the sequence is not Cauchy, it is enough to find one positive tolerance that fails for every proposed \(N\). Take \(\varepsilon=1\). For any positive integer \(N\), choose \(m\geq N\) even and \(n\geq N\) odd. Such indices exist because even and odd integers occur arbitrarily far out. Then \(a_m=1\) and \(a_n=-1\), so

|a_m-a_n|=|1-(-1)|=2\geq1.

Thus no matter how far out the tail begins, it contains two terms whose difference is not less than \(1\). The defining condition fails, so the sequence is not Cauchy and, by the criterion, cannot converge.

Worked Example: Harmonic Partial Sums Are Not Cauchy

Let \(H_n=\sum_{k=1}^{n}1/k\). For every positive integer \(N\), compare \(H_{2N}\) and \(H_N\):

H_{2N}-H_N =\sum_{k=N+1}^{2N}\frac{1}{k} \geq\sum_{k=N+1}^{2N}\frac{1}{2N} =\frac{N}{2N} =\frac{1}{2}.

There are exactly \(N\) summands, and each denominator \(k\) is at most \(2N\), so each summand is at least \(1/(2N)\). For any proposed Cauchy index \(N_0\), use the calculation with \(N=N_0\). The indices \(N_0\) and \(2N_0\) are both at least \(N_0\), but \(|H_{2N_0}-H_{N_0}|\geq1/2\). Taking \(\varepsilon=1/4\), the Cauchy condition fails for every \(N_0\). Therefore the sequence of partial sums is not Cauchy and does not converge.

Why the Real Number System Matters

The two directions of the criterion are not equally general. The proof that convergence implies the Cauchy condition uses only the triangle inequality, so it works for sequences in any metric space. The converse relies on the fact that real numbers have suprema and infima for nonempty bounded sets. This is one way the completeness of \(\mathbb R\) enters the argument.

For a general metric space, a sequence is Cauchy when its terms become arbitrarily close to one another. Such a sequence need not converge to a point of the space unless the space is complete. The Cauchy criterion for real sequences is therefore not just a convenient test: it expresses a completeness property of the real numbers.

When applying the criterion, choose the direction that matches the information available. If a candidate limit is already known, convergence is often straightforward to prove. If no candidate is apparent, compare arbitrary terms in the tail instead. To disprove the condition, identify one fixed positive tolerance and show that separated terms can be found beyond every proposed starting index; a single pair of early terms cannot do this.

Check Your Understanding

Use the definition and the proof of the Cauchy criterion to answer each question.

Why does the proof that a convergent sequence is Cauchy use \(\varepsilon/2\) for each of two terms?
In the converse proof, why are the tail suprema nonincreasing and the tail infima nondecreasing?
How does the Cauchy condition imply that the difference between a tail supremum and tail infimum can be made arbitrarily small?
What fixed tolerance proves that \(((-1)^n)\) is not Cauchy?
Which part of the real-sequence proof uses completeness of \(\mathbb R\)?

What You'll Learn