Limits Along Selected Indices
A sequence can fail to converge even though selected terms approach a definite value. For example, its even-indexed terms might approach one number while its odd-indexed terms approach another. The limits obtained by selecting indices in this way are called subsequential limits.
Earlier in this course, we established that every subsequence of a convergent sequence converges to the same limit. We will use that result as needed, but here we focus on a different question: how can we recognize a number that is the limit of some subsequence? The key is to check whether terms of the original sequence lie arbitrarily close to that number at infinitely many indices. A single close term, or even a finite collection of close terms, is not enough.
Subsequential Limits and Infinite Nearness
The indices of a subsequence are strictly increasing, so a subsequence uses infinitely many distinct indices of the original sequence. The following characterization translates the definition into a condition on those original indices.
Proof. First suppose \(L\) is a subsequential limit. Then there is a subsequence \((a_{n_k})\) with \(a_{n_k}\to L\). Let \(\varepsilon>0\). By the definition of convergence, there is a \(K\in\mathbb{N}_0\) such that $$ |a_{n_k}-L|<\varepsilon $$ for every \(k\geq K\). The indices \(n_K,n_{K+1},n_{K+2},\ldots\) are infinitely many distinct indices of the original sequence. Therefore infinitely many original terms lie within \(\varepsilon\) of \(L\).
Conversely, suppose that for every \(\varepsilon>0\), infinitely many indices \(n\) satisfy \(|a_n-L|<\varepsilon\). We construct a subsequence one index at a time. At step \(k\), the set of indices satisfying $$ |a_n-L|<\frac{1}{k+1} $$ is infinite. Every infinite subset of \(\mathbb{N}_0\) is unbounded, so we can choose such an index \(n_k\) larger than the index chosen at the preceding step. At \(k=0\), choose any qualifying index. This gives strictly increasing indices and $$ |a_{n_k}-L|<\frac{1}{k+1} $$ for every \(k\in\mathbb{N}_0\).
To verify convergence, let \(\varepsilon>0\). Choose \(K\in\mathbb{N}_0\) so that \(1/(K+1)<\varepsilon\). For every \(k\geq K\), we have $$ |a_{n_k}-L|<\frac{1}{k+1}\leq\frac{1}{K+1}<\varepsilon. $$ Thus \(a_{n_k}\to L\), and \(L\) is a subsequential limit. This proves both directions. \(\square\)
The criterion says “infinitely many,” not “eventually all.” To converge to \(L\), a sequence must eventually stay within every positive tolerance of \(L\). To have \(L\) as just one subsequential limit, it is enough that infinitely many terms come within every such tolerance; the other terms may behave very differently.
Worked Examples: Finding and Excluding Subsequential Limits
Worked Example: Two Limits from Alternating Terms
Define $$ a_n=(-1)^n+\frac{1}{n+1}. $$ For even indices \(n=2k\), the terms are $$ a_{2k}=1+\frac{1}{2k+1}. $$ Since \(1/(2k+1)\to0\), the subsequence \((a_{2k})\) converges to \(1\). For odd indices \(n=2k+1\), $$ a_{2k+1}=-1+\frac{1}{2k+2}, $$ which converges to \(-1\). Thus both \(1\) and \(-1\) are subsequential limits.
There are no other subsequential limits. Let \(L\) be distinct from both \(1\) and \(-1\), and set $$ d=\min\{|L-1|,|L+1|\}. $$ Because \(L\) differs from both numbers, \(d>0\). For every \(n\), the value \((-1)^n\) is either \(1\) or \(-1\), so $$ |(-1)^n-L|\geq d. $$ Choose \(N\) so large that \(1/(n+1)<d/2\) whenever \(n\geq N\). The reverse triangle inequality then gives, for \(n\geq N\), $$ |a_n-L| =\left|(-1)^n-L+\frac{1}{n+1}\right| \geq |(-1)^n-L|-\frac{1}{n+1} >d-\frac d2 =\frac d2. $$ Only the finitely many indices \(n<N\) can have \(|a_n-L|<d/2\). By the Infinite Nearness Criterion, \(L\) is not a subsequential limit.
Worked Example: A Value That Occurs Only at Square Indices
Define \(b_n=1\) when \(n=m^2\) for some \(m\in\mathbb{N}_0\), and define \(b_n=0\) otherwise. The indices \(0,1,4,9,\ldots\) are square indices, so the corresponding terms form a constant subsequence equal to \(1\). There are also infinitely many nonsquare indices: for every integer \(m\geq1\), $$ m^2<m^2+1<(m+1)^2, $$ so \(m^2+1\) is not a square. The terms at these indices form a constant subsequence equal to \(0\). Therefore \(0\) and \(1\) are subsequential limits.
To exclude every other number, let \(L\) differ from \(0\) and \(1\), and set \(d=\min\{|L|,|L-1|\}>0\). Every term \(b_n\) is either \(0\) or \(1\), so \(|b_n-L|\geq d\) at every index. In particular, there are no indices with \(|b_n-L|<d/2\). The Infinite Nearness Criterion shows that \(L\) is not a subsequential limit. Thus the only subsequential limits are \(0\) and \(1\).
Worked Example: A Sequence with No Finite Subsequential Limit
Consider \(c_n=n\). Fix any real number \(L\). If \(|c_n-L|<1\), then $$ L-1<n<L+1. $$ There are only finitely many nonnegative integers in this bounded interval, so only finitely many indices \(n\) can satisfy \(|c_n-L|<1\). The Infinite Nearness Criterion fails for \(\varepsilon=1\), no matter which real number \(L\) is chosen. Hence \((c_n)\) has no finite subsequential limit.
This example also warns against assuming that every sequence has a subsequential limit in \(\mathbb{R}\). A subsequential limit must be a real number, and the terms may instead grow without bound. The criterion identifies candidates when they exist; it does not by itself guarantee that one exists.
Eventual Bounds Restrict Subsequential Limits
A sequence may not remain in a fixed interval at every index, but a bound that holds eventually still restricts all of its subsequential limits. The finitely many terms before the bound begins cannot provide infinitely many indices near a candidate limit.
Proof. Suppose, for contradiction, that \(L\) is a subsequential limit and \(L<A\). Set \(\varepsilon=(A-L)/2\), which is positive. If \(|a_n-L|<\varepsilon\), then $$ a_n<L+\varepsilon=\frac{A+L}{2}<A. $$ But \(a_n\geq A\) for every \(n\geq N\). Therefore any index satisfying \(|a_n-L|<\varepsilon\) must be among the finitely many indices \(n<N\). This contradicts the Infinite Nearness Criterion, which requires infinitely many such indices.
Now suppose instead that \(L>B\), and set \(\varepsilon=(L-B)/2>0\). If \(|a_n-L|<\varepsilon\), then $$ a_n>L-\varepsilon=\frac{L+B}{2}>B. $$ For \(n\geq N\), this contradicts \(a_n\leq B\), so again only the finitely many indices \(n<N\) could satisfy the inequality. The Infinite Nearness Criterion gives the same contradiction. Neither \(L<A\) nor \(L>B\) is possible, so \(A\leq L\leq B\). \(\square\)
The conclusion concerns subsequential limits, not the terms at every index. A finite number of terms may lie outside \([A,B]\), and that does not affect the result. For instance, if a sequence is eventually nonnegative, every subsequential limit must be nonnegative: if \(L<0\), take \(\varepsilon=-L/2>0\). Then \(|a_n-L|<\varepsilon\) implies \(a_n<L/2<0\), so only finitely many indices can satisfy this, contradicting the Infinite Nearness Criterion. More generally, whenever an eventual interval bound is known, no subsequential limit can lie strictly outside that interval.
Why Infinite Nearness Is the Right Test
When testing a proposed subsequential limit, ask whether every neighborhood of the candidate contains terms at infinitely many indices. If one tolerance leaves only finitely many eligible indices, the candidate cannot be a subsequential limit. If every tolerance has infinitely many eligible indices, the theorem provides a systematic construction: at step \(k\), choose a later index whose term is within \(1/(k+1)\) of the candidate.
A common pitfall is to count repeated values rather than nearby terms. A candidate need not occur exactly as a term even once. For example, the sequence \(1/(n+1)\) never equals \(0\), but it has infinitely many terms within every positive distance of \(0\); the reciprocal sequence converges to \(0\), so \(0\) is a subsequential limit. Conversely, having infinitely many terms within one fixed distance is insufficient. The required condition must hold for every positive \(\varepsilon\), however small.
Subsequential limits also provide a useful way to detect failure of convergence. If two subsequences converge to distinct real numbers, the original sequence cannot converge, by the Distinct Subsequence Limits Obstruct Convergence theorem established earlier. The examples above illustrate the underlying pattern: different subsequences may approach different values, or the sequence may have no finite subsequential limits at all. The next step in studying this relationship is to examine more closely what happens to subsequences when the original sequence does converge.
Check Your Understanding
Use the Infinite Nearness Criterion and the eventual-bounds result to answer the following questions.
- What condition on the indices within \(\varepsilon\) of \(L\) characterizes \(L\) as a subsequential limit?
- Why does a subsequence converging to \(L\) give infinitely many original indices within every positive tolerance of \(L\)?
- For the sequence \(b_n\) that equals \(1\) at square indices and \(0\) elsewhere, why are both \(0\) and \(1\) subsequential limits?
- Can a number that occurs as a term only once be a subsequential limit? Explain what the criterion requires instead.
- If \(A\leq a_n\leq B\) for all sufficiently large \(n\), where must every subsequential limit lie?