Cauchy Sequences

When a Sequence Is Called Cauchy

The previous tutorial introduced the Cauchy property: sufficiently late terms of a sequence must be close to one another, uniformly over every pair of indices in the tail. A real sequence with this property is called a Cauchy sequence. The name is useful because it lets us discuss not just the condition, but also how it behaves when sequences are selected, modified, or combined.

Definition (Cauchy Sequence): A real sequence \((a_n)_{n=0}^{\infty}\) is a Cauchy sequence if for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}_0\) such that, whenever \(m,n\geq N\), \(|a_m-a_n|<\varepsilon\).

The same index \(N\) must work for every pair \(m,n\) in the tail. In particular, the condition is stronger than asking adjacent terms to get close: two indices in the tail may be arbitrarily far apart. The previous tutorial established that every convergent real sequence is Cauchy, and that a real sequence with the Cauchy property is bounded. We will use these facts as tools rather than prove them again.

Direct Estimates for the Cauchy Condition

To verify that a sequence is Cauchy directly, begin with arbitrary \(m,n\geq N\) and bound \(|a_m-a_n|\) by an expression that becomes small as \(N\) grows. The estimate must not depend on how far apart \(m\) and \(n\) are.

Worked Example: A Sequence Approaching a Fixed Number

Let \(a_n=3+\frac{1}{n+2}\). For \(m,n\geq N\), the triangle inequality gives

|a_m-a_n| =\left|\frac{1}{m+2}-\frac{1}{n+2}\right| \leq \frac{1}{m+2}+\frac{1}{n+2} \leq \frac{2}{N+2}.

Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) so that \(2/(N+2)<\varepsilon\). Then \(|a_m-a_n|<\varepsilon\) for every \(m,n\geq N\), so \((a_n)\) is Cauchy. The estimate is uniform across the tail: it covers every pair of indices, not just consecutive ones.

A sequence can fail the condition even if the difference between consecutive terms becomes small. The next example checks the full pairwise requirement rather than relying on adjacent differences.

Worked Example: Small Steps Do Not Ensure the Cauchy Property

Let \(b_n=\sqrt{n}\), for \(n\in\mathbb{N}_0\). Its successive differences become small, since for \(n\geq1\),

|b_{n+1}-b_n| =\sqrt{n+1}-\sqrt n =\frac{1}{\sqrt{n+1}+\sqrt n} \leq\frac{1}{2\sqrt n}.

However, \((b_n)\) is not Cauchy. Use the tolerance \(\varepsilon=1\). For any proposed \(N\), let \(r=\max\{N,1\}\), \(n=r\), and \(m=4r\). Both indices are at least \(N\), but

|b_m-b_n| =\sqrt{4r}-\sqrt r =2\sqrt r-\sqrt r =\sqrt r \geq1.

Thus no proposed \(N\) makes every pair of terms in the tail less than \(1\) apart. This example shows why a Cauchy estimate must control all pairs, even when the steps from one term to the next shrink.

Subsequences and Finite Changes

Selecting terms from a Cauchy sequence cannot destroy its tail closeness. A subsequence eventually selects only terms from any chosen tail of the original sequence.

Theorem (Every Subsequence of a Cauchy Sequence Is Cauchy): If \((a_n)\) is a Cauchy sequence and \((a_{n_k})\) is a subsequence, then \((a_{n_k})\) is Cauchy.

Proof. Let \(\varepsilon>0\). Since \((a_n)\) is Cauchy, there is an \(N\in\mathbb{N}_0\) such that \(|a_i-a_j|<\varepsilon\) whenever \(i,j\geq N\). The indices of a subsequence are strictly increasing, so they eventually exceed \(N\). Choose \(K\) such that \(n_k\geq N\) for every \(k\geq K\). If \(p,q\geq K\), then \(n_p,n_q\geq N\), and therefore

|a_{n_p}-a_{n_q}|<\varepsilon.

The same \(K\) works for every pair \(p,q\geq K\). Hence the subsequence is Cauchy. \(\square\)

Changing finitely many terms also leaves the Cauchy property unchanged. The definition only constrains a sufficiently late tail, so any finite collection of initial terms can be excluded by moving the starting index farther out.

Theorem (Finite Changes Preserve the Cauchy Property): Suppose two real sequences \((a_n)\) and \((b_n)\) agree for every \(n\geq M\), for some \(M\in\mathbb{N}_0\). Then \((a_n)\) is Cauchy if and only if \((b_n)\) is Cauchy.

Proof. Suppose first that \((a_n)\) is Cauchy, and let \(\varepsilon>0\). Choose \(N_1\) such that \(|a_m-a_n|<\varepsilon\) whenever \(m,n\geq N_1\). Set \(N=\max\{N_1,M\}\). For \(m,n\geq N\), both indices are at least \(M\), so \(a_m=b_m\) and \(a_n=b_n\). Thus

|b_m-b_n|=|a_m-a_n|<\varepsilon.

Therefore \((b_n)\) is Cauchy. Interchanging the roles of the two sequences proves the converse. \(\square\)

Worked Example: Altering Initial Terms

Define \(x_n=1/(n+1)\), and define \(y_0=40\), \(y_1=-7\), and \(y_n=x_n\) for \(n\geq2\). First, \((x_n)\) is Cauchy: if \(m,n\geq N\), then

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

For any \(\varepsilon>0\), choose \(N\geq2\) large enough that \(2/(N+1)<\varepsilon\). This proves the Cauchy condition for \((x_n)\). The sequences \(x_n\) and \(y_n\) agree for all \(n\geq2\), so finite changes preserve the Cauchy property and \((y_n)\) is Cauchy as well. The two unusually large initial values do not affect the tail condition.

Combining Cauchy Sequences

Cauchy sequences are stable under addition and scalar multiplication. For addition, compare two sums by separating their differences. For a scalar multiple, the scalar simply scales the distance between the terms.

Theorem (Sums and Scalar Multiples of Cauchy Sequences): If \((a_n)\) and \((b_n)\) are Cauchy sequences and \(c\in\mathbb{R}\), then \((a_n+b_n)\) and \((ca_n)\) are Cauchy sequences.

Proof. Let \(\varepsilon>0\). Since \((a_n)\) and \((b_n)\) are Cauchy, there are indices \(N_a,N_b\) such that \(|a_m-a_n|<\varepsilon/2\) for \(m,n\geq N_a\), and \(|b_m-b_n|<\varepsilon/2\) for \(m,n\geq N_b\). For \(m,n\geq \max\{N_a,N_b\}\),

|(a_m+b_m)-(a_n+b_n)| \leq |a_m-a_n|+|b_m-b_n| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

Thus \((a_n+b_n)\) is Cauchy. If \(c=0\), then \(ca_n=0\) for every \(n\), so the scalar multiple is Cauchy. If \(c\neq0\), apply the Cauchy condition for \((a_n)\) with tolerance \(\varepsilon/|c|\). For sufficiently large \(m,n\), this gives

|ca_m-ca_n| =|c|\,|a_m-a_n| <|c|\frac{\varepsilon}{|c|} =\varepsilon.

Therefore \((ca_n)\) is Cauchy in all cases. \(\square\)

Products require one additional ingredient. The difference of two products contains both a difference in the first sequence and a difference in the second. Boundedness controls the factors multiplying those differences. By the theorem from the previous tutorial, every Cauchy sequence of real numbers is bounded.

Theorem (Products of Cauchy Sequences): If \((a_n)\) and \((b_n)\) are Cauchy sequences, then \((a_nb_n)\) is Cauchy.

|a_mb_m-a_nb_n| \leq |a_m|\,|b_m-b_n|+|b_n|\,|a_m-a_n| <K\frac{\varepsilon}{2K}+K\frac{\varepsilon}{2K} =\varepsilon.

This holds for every \(m,n\geq N\), so the product sequence is Cauchy. \(\square\)

Worked Example: A Product of Cauchy Sequences

Let \(x_n=1+\frac{1}{n+1}\) and \(y_n=2-\frac{1}{n+2}\). For \(m,n\geq N\),

|x_m-x_n| =\left|\frac{1}{m+1}-\frac{1}{n+1}\right| \leq\frac{2}{N+1}, \qquad |y_m-y_n| =\left|\frac{1}{m+2}-\frac{1}{n+2}\right| \leq\frac{2}{N+2}.

Both bounds can be made smaller than any prescribed positive tolerance by choosing \(N\) large enough. Thus both input sequences are Cauchy. The product theorem now shows that \((x_ny_n)\) is Cauchy. To see the product explicitly,

x_ny_n =\left(1+\frac{1}{n+1}\right)\left(2-\frac{1}{n+2}\right) =2+\frac{2}{n+1}-\frac{1}{n+2} -\frac{1}{(n+1)(n+2)}.

The theorem avoids having to estimate the difference of these four terms separately: boundedness of the factors and the Cauchy estimates for each input sequence already control the difference between any two products.

Why These Closure Properties Matter

The definition gives a test based only on terms of the sequence, without requiring a proposed limit. The results above make that test practical: subsequences remain Cauchy, finite initial changes are irrelevant, and sums, scalar multiples, and products preserve the property. When analyzing an expression built from sequences, these closure results can replace a fresh pairwise estimate for every term.

Keep the scope of the Cauchy condition in mind. It is not enough to find one pair of close terms, or to show that consecutive terms become close. For a positive result, a single index must control every pair in the tail. For a negative result, it is enough to find one tolerance such that every proposed tail contains a pair at least that far apart.

Check Your Understanding

Use the definition and the results in this tutorial to answer the following questions.

What must a single index \(N\) guarantee in the definition of a Cauchy sequence?
Why does every subsequence of a Cauchy sequence satisfy the Cauchy condition?
How does choosing \(N\geq M\) help prove that finite changes preserve the Cauchy property?
Where is boundedness used in the proof that the product of two Cauchy sequences is Cauchy?
Why do small successive differences alone fail to establish that a sequence is Cauchy?

What You'll Learn