From Pairwise Closeness to a Limit
The Cauchy criterion says that a real sequence converges exactly when its late terms become uniformly close to one another. In the previous tutorial, we used this criterion to test sequences without first knowing their limits. Here we prove the direction that makes this possible: every Cauchy sequence of real numbers converges to a finite real number.
The key step is to find a candidate limit. A Cauchy sequence is bounded, and the Bolzano–Weierstrass Theorem then gives it a convergent subsequence. The Cauchy property ensures that the rest of the sequence cannot stay far from that subsequence’s limit. We will first make this last step quantitative, then use it to prove the criterion.
The Cauchy property guarantees that at least one such index can be chosen for each positive tolerance. A modulus records those choices. Likewise, if a subsequence \((a_{n_k})\) converges to \(L\), a convergence modulus for that subsequence is a function \(J:(0,\infty)\to\mathbb{N}_0\) such that \(k\geq J(\varepsilon)\) implies \(|a_{n_k}-L|<\varepsilon\). We use the standard indexing \(n_0<n_1<n_2<\cdots\); in particular, \(n_k\geq k\), as established earlier in the course.
A Quantitative Bridge from a Subsequence
The following result supplies both the logical bridge and an explicit way to choose a convergence index. It does not require a formula for the limit: it only requires a convergent subsequence and a Cauchy modulus for the original sequence.
Proof. Fix \(\varepsilon>0\), and let \(j=j(\varepsilon)\) be as in the statement. Since \(j\geq J(\varepsilon/2)\), the subsequence convergence modulus gives $$ |a_{n_j}-L|<\frac{\varepsilon}{2}. $$ Also, \(j\geq M(\varepsilon/2)\). Because subsequence indices satisfy \(n_j\geq j\), we have \(n_j\geq M(\varepsilon/2)\). Set \(N=N(\varepsilon)\). If \(n\geq N\), then \(n\geq M(\varepsilon/2)\), and \(n_j\geq M(\varepsilon/2)\) as well. The Cauchy modulus therefore gives $$ |a_n-a_{n_j}|<\frac{\varepsilon}{2}. $$ By the triangle inequality, $$ |a_n-L|\leq |a_n-a_{n_j}|+|a_{n_j}-L|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ This holds for every \(n\geq N(\varepsilon)\), so \(N\) is a convergence modulus for \((a_n)\) and \(a_n\to L\). \(\square\)
The choice \(n_j\) is important: the Cauchy estimate compares terms whose original sequence indices are at least \(M(\varepsilon/2)\). A large subsequence index \(j\) ensures that the selected term \(a_{n_j}\) is far enough out in the original sequence as well. The two error allowances, each \(\varepsilon/2\), then combine to give the required total error \(\varepsilon\).
Worked Example: A Cauchy Sequence with an Explicit Subsequence Limit
Define \(b_n=4+1/(n+1)\) for \(n\in\mathbb{N}_0\). First, for any \(m,n\geq N\), the triangle inequality gives $$ |b_m-b_n| =\left|\frac{1}{m+1}-\frac{1}{n+1}\right| \leq \frac{1}{m+1}+\frac{1}{n+1} \leq \frac{2}{N+1}. $$ For any \(\varepsilon>0\), choosing \(N=\lfloor 2/\varepsilon\rfloor+1\) makes \(N+1>2/\varepsilon\), and hence \(2/(N+1)<\varepsilon\). Thus \(M(\varepsilon)=\lfloor 2/\varepsilon\rfloor+1\) is a Cauchy modulus.
The even-indexed subsequence is \(b_{2k}=4+1/(2k+1)\), which converges to \(4\). Indeed, for any \(\delta>0\), if \(k\geq\lfloor 1/\delta\rfloor+1\), then \(2k+1>1/\delta\), so \(|b_{2k}-4|=1/(2k+1)<\delta\). The quantitative bridge theorem now gives convergence of the full sequence to \(4\). In this example the limit is also apparent from the formula, but the theorem’s method only needs the Cauchy estimate and the convergent subsequence.
Proof of the Cauchy Criterion
Proof. If \((a_n)\) converges, then it has the Cauchy property by the theorem Every Convergent Sequence Has the Cauchy Property, established earlier in the course. It remains to prove the converse.
Suppose that \((a_n)\) is Cauchy. By the theorem A Sequence with the Cauchy Property Is Bounded, it is bounded. The Bolzano–Weierstrass Theorem states that every bounded real sequence has a convergent subsequence. Thus there are strictly increasing indices \(n_0<n_1<n_2<\cdots\) and a real number \(L\) such that \(a_{n_k}\to L\).
The sequence has a Cauchy modulus by the definition of the Cauchy property, and the convergent subsequence has a convergence modulus by the definition of convergence. Apply A Cauchy Modulus and a Convergent Subsequence Give a Convergence Modulus. It follows that \(a_n\to L\). In particular, the limit is a finite real number, as required. This proves the reverse direction and hence the criterion. \(\square\)
The established theorem A Cauchy Sequence with a Convergent Subsequence Converges gives the same logical bridge without keeping track of indices quantitatively. The modulus version above makes the mechanism explicit: choose one subsequence term close to \(L\) and sufficiently far out, then compare every sufficiently late term of the full sequence with it.
Worked Example: Telescoping Terms and the Cauchy Criterion
For \(n\in\mathbb{N}_0\), let $$ c_n=\sum_{j=1}^{n+1}\frac{1}{j(j+1)}. $$ The identity \(1/[j(j+1)]=1/j-1/(j+1)\) gives $$ c_n=1-\frac{1}{n+2}. $$ For \(m,n\geq N\), it follows that $$ |c_m-c_n| =\left|\frac{1}{n+2}-\frac{1}{m+2}\right| \leq \frac{1}{n+2}+\frac{1}{m+2} \leq \frac{2}{N+2}. $$ Given \(\varepsilon>0\), choose \(N=\lfloor 2/\varepsilon\rfloor+1\). Then \(N+2>2/\varepsilon\), so \(2/(N+2)<\varepsilon\). The sequence is Cauchy and therefore converges by the criterion. Here its limit can also be identified directly: \(c_n=1-1/(n+2)\to1\).
Worked Example: Why a Bounded Sequence Alone Is Not Enough
Consider \(d_n=(-1)^n/2\). It is bounded, since \(|d_n|=1/2\) for every \(n\), and its even-indexed subsequence is constantly \(1/2\). But the full sequence is not Cauchy. Given any \(N\), choose an even \(m\geq N\) and an odd \(n\geq N\). Then $$ |d_m-d_n|=\left|\frac12-\left(-\frac12\right)\right|=1. $$ Thus every tail contains a pair separated by \(1\). The convergent subsequence alone cannot force the whole sequence to converge; the Cauchy property is what controls the terms not on that subsequence.
Where Completeness Enters
The proof uses two ingredients with different roles. The Cauchy property gives boundedness and ensures that sufficiently late terms are close to one another. Bolzano–Weierstrass supplies a convergent subsequence and hence a real candidate \(L\). The quantitative bridge then transfers convergence from that subsequence to the entire sequence. This is the point at which the special structure of \(\mathbb{R}\) matters: the subsequence is guaranteed to have a limit in \(\mathbb{R}\).
A common incomplete argument says, “The sequence is Cauchy, so it has a limit,” and stops there. That sentence is the conclusion of the criterion, not a proof of it. To establish the conclusion, one must produce a candidate limit or invoke a previously proved result that does so. Here boundedness and Bolzano–Weierstrass produce the candidate, and the Cauchy estimate verifies that it is the limit of the whole sequence.
Another pitfall is to establish convergence only along a subsequence and assume that this settles convergence of the original sequence. The alternating example shows why that is insufficient. In the criterion’s proof, the convergent subsequence identifies the candidate, while the Cauchy condition controls all late terms, including those whose indices are not among the selected subsequence indices.
Check Your Understanding
Use the proof strategy and quantitative estimate to answer the following questions.
- Which theorem provides a convergent subsequence after boundedness of a Cauchy sequence is known?
- In the quantitative bridge theorem, why is \(n_j\geq M(\varepsilon/2)\) needed as well as \(j\geq J(\varepsilon/2)\)?
- How does the triangle inequality combine the two error bounds in the proof of the bridge theorem?
- Why does the proof of the reverse direction need both a convergent subsequence and the Cauchy property?
- What fails if one knows only that a bounded sequence has a convergent subsequence?