Let a Subsequence Test the Claim
In “The Contradiction Strategy,” a temporary assumption was used to derive a precise incompatibility. A related strategy for sequence proofs is to pass to a subsequence: keep only selected terms, in their original order, and use those terms to test a claim about the whole sequence. A well-chosen subsequence can reveal persistent behavior that the full sequence obscures.
The key is that a subsequence does not rearrange the sequence. It selects indices that keep increasing. This simple restriction makes subsequences useful in two opposite ways: if a sequence converges, every subsequence must have the same limit; if convergence fails, one can select terms that remain a fixed distance from the proposed limit.
Earlier in this course, the theorem “Subsequences Preserve Limits” established that if \(x_n\to L\), then every subsequence \(x_{n_k}\) also converges to \(L\). Therefore, finding two subsequences with different limits is enough to rule out convergence of the original sequence. More generally, a subsequence that stays away from \(L\) is evidence against the claim that \(x_n\to L\).
Extracting a Subsequence That Stays Away
The negation of convergence gives exactly the information needed to construct a useful subsequence. The statement \(x_n\to L\) says that for every \(\varepsilon>0\), all sufficiently late terms lie within \(\varepsilon\) of \(L\). Its negation says that some fixed positive distance is missed arbitrarily late: no matter how far out we go, a term remains at least that distance from \(L\).
Proof. First, suppose such an \(\varepsilon\) and subsequence exist. If \(x_n\to L\), the theorem “Subsequences Preserve Limits” would imply \(x_{n_k}\to L\). But convergence of \(x_{n_k}\) to \(L\), applied with tolerance \(\varepsilon\), would give an index \(K\) such that \(|x_{n_k}-L|<\varepsilon\) for every \(k\geq K\). This contradicts \(|x_{n_k}-L|\geq\varepsilon\) for every \(k\). So \(x_n\) cannot converge to \(L\).
Conversely, suppose \(x_n\) fails to converge to \(L\). Negating the definition of convergence gives an \(\varepsilon>0\) such that for every positive integer \(N\), there is an \(n\geq N\) with \(|x_n-L|\geq\varepsilon\). Choose such an index \(n_1\) for \(N=1\). Once \(n_k\) has been chosen, apply the same statement with \(N=n_k+1\) to choose \(n_{k+1}>n_k\) satisfying \(|x_{n_{k+1}}-L|\geq\varepsilon\). This constructs a strictly increasing sequence of indices, and every selected term stays at least \(\varepsilon\) away from \(L\). Thus there is a subsequence of the required form. \(\square\)
The recursive choice matters: choosing a distant term repeatedly without ensuring that its index is larger would not necessarily produce a subsequence. The theorem does not say that every subsequence stays away from \(L\). It says that when convergence fails, one can deliberately select a subsequence that does.
Worked Example: Detecting Oscillation with Two Subsequences
For positive integers \(n\), define \(x_n=(-1)^n+1/n\). The even-indexed terms are
so \(x_{2k}\to1\). The odd-indexed terms are
so \(x_{2k-1}\to-1\). If the original sequence converged to some \(L\), both subsequences would converge to \(L\), by “Subsequences Preserve Limits.” The uniqueness of sequence limits would then force \(L=1\) and \(L=-1\), which is impossible. Hence \((x_n)\) does not converge.
The two subsequences identify the obstruction: the even terms approach one value while the odd terms approach another. The argument uses only selected terms and the established limit theorems; it does not require a separate estimate of how far every term is from a proposed limit.
Worked Example: A Limit of Squares Forces a Limit of the Terms
Suppose a real sequence satisfies \(x_n^2\to0\). We prove that \(x_n\to0\) using the subsequence obstruction strategy. Assume instead that \(x_n\) does not converge to \(0\). The theorem gives an \(\varepsilon>0\) and a subsequence \((x_{n_k})\) such that \(|x_{n_k}|\geq\varepsilon\) for every \(k\). Squaring this inequality gives
for every \(k\). On the other hand, \(x_n^2\to0\), so “Subsequences Preserve Limits” implies \(x_{n_k}^2\to0\). Convergence to zero, with tolerance \(\varepsilon^2>0\), requires \(x_{n_k}^2<\varepsilon^2\) for all sufficiently large \(k\). This contradicts the displayed lower bound. Therefore \(x_n\to0\).
The subsequence makes the negation of the desired conclusion quantitative: its terms have absolute value bounded below by one fixed positive number. Squaring transfers that lower bound to a subsequence of the squares, where it conflicts with the assumed limit.
A Criterion Using Further Subsequences
Sometimes a proof does not directly show that every term is close to \(L\). It may be easier to show that, from any subsequence, one can select a further subsequence that does converge to \(L\). Surprisingly, that condition alone forces the original sequence to converge to \(L\). The reason is that a subsequence witnessing failure of convergence cannot have any further subsequence converging to \(L\).
Proof. Suppose first that \(x_n\to L\). Every subsequence \(x_{n_k}\) converges to \(L\), by “Subsequences Preserve Limits.” In particular, it has a further subsequence converging to \(L\): the subsequence itself, obtained by taking its indices \(1,2,3,\ldots\).
For the converse, suppose every subsequence has a further subsequence converging to \(L\), but \(x_n\) does not converge to \(L\). By the “Subsequence Obstruction to Convergence” theorem, there are an \(\varepsilon>0\) and a subsequence \((x_{n_k})\) satisfying \(|x_{n_k}-L|\geq\varepsilon\) for every \(k\). By the assumed property, this subsequence has a further subsequence converging to \(L\). But every term of that further subsequence still has distance at least \(\varepsilon\) from \(L\), so it cannot converge to \(L\). This is a contradiction. Hence \(x_n\to L\). \(\square\)
The criterion is useful when a sequence has complicated behavior but all its possible subsequential behavior can be controlled. The quantifiers are essential: it says every subsequence has a further subsequence converging to \(L\). Finding just one subsequence that converges to \(L\) does not establish convergence of the original sequence.
Worked Example: Using the Further-Subsequence Criterion
Define \(x_n=\sin(n^2)/(n+1)\). We show that \(x_n\to0\) by verifying the condition in the Further-Subsequence Criterion. Consider any subsequence \((x_{n_k})\). Since \(n_1<n_2<\cdots\) are positive integers, \(n_k\geq k\). Also, \(|\sin(n_k^2)|\leq1\). Therefore
As \(k\to\infty\), the right-hand side tends to \(0\), so this arbitrary subsequence itself converges to \(0\). In particular, it has a further subsequence converging to \(0\). Since the same reasoning applies to every subsequence, the Further-Subsequence Criterion gives \(x_n\to0\).
The estimate uses the increasing-index condition directly: selected indices satisfy \(n_k\geq k\), so the denominator still grows along every subsequence. A subsequence may skip many terms, but its \(k\)-th index cannot be smaller than \(k\).
Choosing the Right Subsequence
A subsequence argument is most effective when the index choice is tied to the claim being tested. To show that a sequence does not converge to \(L\), select indices where the distance from \(L\) is at least a fixed \(\varepsilon\). To rule out convergence altogether, two subsequences with different limits are often convenient. To prove convergence using the further-subsequence criterion, start with an arbitrary subsequence and show that it must contain a further subsequence converging to the proposed limit.
For a proposed limit \(L\), note whether the goal is to prove convergence to \(L\) or to disprove it.
If convergence fails, obtain a fixed \(\varepsilon>0\) and arbitrarily late indices whose terms are at least \(\varepsilon\) from \(L\).
Make each new index larger than the previous one, so the selected terms really form a subsequence.
Compare the behavior of the selected terms with the assumed limit, or apply the further-subsequence criterion when it is available.
A common mistake is to infer that a sequence converges because one subsequence converges. For example, the even-indexed terms of the oscillating sequence in the first example converge to \(1\), but the odd-indexed terms converge to \(-1\). A convergent subsequence describes selected terms, not necessarily the full sequence. The valid implication goes the other way: convergence of the full sequence forces every subsequence to have the same limit.
Another important distinction is between a subsequence and a rearrangement. The indices must increase strictly. If the proof selects terms in a different order, it has changed the sequence in a way not covered by “Subsequences Preserve Limits.” Keeping the order intact is what makes the subsequence strategy reliable.
Check Your Understanding
Use the definition of a subsequence and the two convergence criteria to answer the following questions.
- Why must the indices in a subsequence be strictly increasing?
- What does failure of \(x_n\to L\) guarantee about some fixed positive distance from \(L\)?
- How can two subsequences with different limits show that the original sequence does not converge?
- In the Further-Subsequence Criterion, why is it not enough to find just one subsequence converging to \(L\)?
- In the example \(x_n=\sin(n^2)/(n+1)\), why is \(n_k\geq k\) for every subsequence?