When a Property Holds Infinitely Often
A convergence proof controls every term in a sufficiently late tail. A subsequence proof can focus on a different pattern: a property that occurs infinitely often, even if it does not hold for all sufficiently large indices. The key move is to select just the terms where the property holds. The result is a subsequence, and if the original sequence converges, the theorem Subsequences of a Convergent Sequence says that this selected sequence has the same limit.
The construction is available from the earlier theorem Constructing a Subsequence from Infinitely Many Eligible Indices: infinitely many eligible indices can be listed in strictly increasing order. This lets us turn an “infinitely often” statement into a sequence of terms to which convergence can be applied. The technique is useful both for proving restrictions on a limit and for showing that a sequence cannot converge.
The selected terms need not form a consecutive block, and they need not include every term. Their value is that they preserve the original limit while retaining a property that may occur only intermittently.
Recurring Bounds Constrain the Limit
Proof. First suppose \(a_n\geq c\) for infinitely many indices. By Constructing a Subsequence from Infinitely Many Eligible Indices, choose strictly increasing indices \(n_0<n_1<n_2<\cdots\) such that \(a_{n_k}\geq c\) for every \(k\). The sequence \((a_{n_k})\) is a subsequence of \((a_n)\), so \(a_{n_k}\to L\) by Subsequences of a Convergent Sequence.
Suppose, contrary to the desired conclusion, that \(L<c\). Apply the definition of convergence of \(a_{n_k}\) with \(\varepsilon=(c-L)/2>0\). For all sufficiently large \(k\), it gives $$ |a_{n_k}-L|<\frac{c-L}{2}. $$ In particular, $$ a_{n_k}<L+\frac{c-L}{2}=\frac{L+c}{2}<c. $$ This contradicts \(a_{n_k}\geq c\) for every \(k\). Therefore \(L\geq c\).
Now suppose \(a_n\leq c\) for infinitely many indices. Select a subsequence \((a_{m_k})\) with \(a_{m_k}\leq c\) for every \(k\). It also converges to \(L\). If \(L>c\), convergence with \(\varepsilon=(L-c)/2>0\) implies, for all sufficiently large \(k\), $$ a_{m_k}>L-\frac{L-c}{2}=\frac{L+c}{2}>c. $$ This contradicts \(a_{m_k}\leq c\). Hence \(L\leq c\), proving both claims. \(\square\)
The inequalities in the proof illustrate the central logic. A recurring lower bound rules out a limit strictly below it; a recurring upper bound rules out a limit strictly above it. The property need not hold at every index, because it holds at every index of the selected subsequence.
Worked Example: A Limit from Terms on Both Sides
Define $$ a_n=3+\frac{(-1)^n}{n+2}. $$ For every \(n\in\mathbb{N}_0\), $$ |a_n-3|=\frac{1}{n+2}\leq\frac{1}{n+1}, $$ and \(1/(n+1)\to0\). The Squeeze Theorem therefore gives \(a_n\to3\).
For every even index \(n=2k\), we have \((-1)^{2k}=1\), so $$ a_{2k}=3+\frac{1}{2k+2}>3. $$ Thus \(a_n\geq3\) for infinitely many indices, and the recurring lower-bound result implies that the limit is at least \(3\). For every odd index \(n=2k+1\), \((-1)^{2k+1}=-1\), so $$ a_{2k+1}=3-\frac{1}{2k+3}<3. $$ The recurring upper-bound result says that the limit is at most \(3\). Together these bounds confirm that the limit is \(3\).
Worked Example: Infinitely Many Terms in a Closed Interval
Suppose \(a_n\to L\), and suppose \(a_n^2\leq16\) for infinitely many indices. For a real number \(x\), the inequality \(x^2\leq16\) is equivalent to \(-4\leq x\leq4\): indeed, \(x^2\leq16\) gives \(|x|\leq4\), which means exactly \(-4\leq x\leq4\).
At each of the infinitely many eligible indices we therefore have both \(a_n\geq-4\) and \(a_n\leq4\). Applying the infinitely often lower-bound result with \(c=-4\) gives \(L\geq-4\). Applying the upper-bound result with \(c=4\) gives \(L\leq4\). Consequently, $$ -4\leq L\leq4. $$ The conclusion concerns the limit, not necessarily every term of the original sequence. Terms outside the interval may occur at other indices without affecting this argument.
Infinitely Often Conditions Can Prove Nonconvergence
The same reasoning gives a way to test a proposed limit. If convergence to \(L\) would force \(L\) to satisfy two incompatible inequalities, then the sequence cannot converge. The recurring conditions can concern different sets of indices: one subsequence may supply a lower bound, while another supplies an upper bound.
Worked Example: Incompatible Recurring Bounds
Consider $$ a_n=(-1)^n\left(2+\frac{1}{n+1}\right). $$ At every even index \(n=2k\), direct substitution gives $$ a_{2k}=2+\frac{1}{2k+1}\geq2. $$ There are infinitely many even indices. If \((a_n)\) converged to \(L\), the infinitely often lower-bound result would imply \(L\geq2\).
At every odd index \(n=2k+1\), substitution instead gives $$ a_{2k+1}=-\left(2+\frac{1}{2k+2}\right)\leq-2. $$ There are infinitely many odd indices as well. Convergence would then imply \(L\leq-2\). No real number can satisfy both \(L\geq2\) and \(L\leq-2\). Therefore the sequence does not converge.
This proof does not need to calculate limits of the even and odd subsequences. It uses only recurring bounds and the fact that any subsequence of a convergent sequence must share the same limit.
Proof. The condition \(a_n\in[A,B]\) means \(A\leq a_n\leq B\). In particular, \(a_n\geq A\) at infinitely many indices, so the infinitely often lower-bound result gives \(L\geq A\). Also \(a_n\leq B\) at infinitely many indices, so the upper-bound result gives \(L\leq B\). Hence \(A\leq L\leq B\), which is precisely \(L\in[A,B]\). \(\square\)
The interval theorem is useful when a condition is naturally stated as membership in a bounded interval. It applies to any closed interval, including the degenerate case \(A=B\). In that case, if infinitely many terms equal \(A\) and the sequence converges, its limit must equal \(A\).
Worked Example: Repeated Terms Determine the Limit
Suppose \(a_n\to L\), and assume that \(a_n=7\) for infinitely many indices. Each of those terms belongs to the interval \([7,7]\). Applying the limit-from-an-interval theorem with \(A=B=7\) gives $$ L\in[7,7], $$ so \(L=7\).
Equivalently, select a subsequence consisting of the terms equal to \(7\). It is the constant sequence \(7,7,7,\ldots\), and it must converge to \(L\) because it is a subsequence of a sequence converging to \(L\). The theorem The Limit of a Constant Sequence then gives \(L=7\). The interval argument packages the same constraint in a form that also handles nonconstant selected terms.
Choosing the Right Subsequence
A useful subsequence proof begins by identifying what must be retained. If the aim is to preserve an inequality, choose indices where that inequality holds. If the aim is to show that convergence to \(L\) is impossible, look for infinitely many terms on each side of a proposed bound, or infinitely many terms that violate a fixed tolerance. The construction should preserve the relevant evidence while the inheritance theorem carries convergence to the selected sequence.
Specify the set of indices where the needed inequality or interval condition holds.
Use the theorem on constructing a subsequence from infinitely many eligible indices to list those indices in increasing order.
If the original sequence converges to \(L\), the selected subsequence also converges to \(L\).
Use the epsilon definition, or the recurring-bounds theorem, to derive the required restriction on \(L\).
A common pitfall is to treat an infinitely often property as though it held eventually. “Infinitely many terms satisfy \(a_n\geq c\)” does not say that all sufficiently late terms satisfy \(a_n\geq c\). The subsequence method avoids that mistake: it uses the inequality only on the selected indices, where it is guaranteed to hold.
A second pitfall is to infer a restriction on the limit from a property that occurs only finitely many times. A finite set of exceptional terms is not enough to construct an infinite subsequence. For example, one term above \(c\) says nothing by itself about whether the limit is above or below \(c\). The infinitely often hypothesis is what supplies arbitrarily large eligible indices.
In practice, subsequences are most effective when the original sequence mixes different behaviors. Instead of estimating every term at once, separate the indices according to a useful condition. If the resulting restrictions on a common limit contradict one another, convergence is impossible; if they are compatible, they can locate or bound the limit.
Check Your Understanding
Use the subsequence method and the results established here to answer the following questions.
- If \(a_n\to L\) and \(a_n\geq c\) for infinitely many indices, what inequality must \(L\) satisfy?
- Why does an infinitely often condition allow a subsequence to be selected, while a condition holding only finitely often does not?
- If \(a_n\to L\) and \(a_n\in[-2,5]\) for infinitely many indices, what can be concluded about \(L\)?
- How can recurring lower and upper bounds be used to prove that a sequence diverges?
- Why is it incorrect to say that an infinitely often inequality holds for every sufficiently large index?