Selecting Terms Without Changing Their Order
A subsequence is formed by selecting terms from a sequence while keeping their original order. This lets us examine behavior that may be hidden when the full sequence is considered all at once. In the previous tutorial, Cluster Points of Sequences, a cluster point was defined as the limit of a subsequence. We now look more closely at how such subsequences are indexed and how their convergence can be verified.
The strict increase of the indices is essential. It ensures that each selected term comes after the one before it and that no original index is used twice. A subsequence does not have to select terms at regular intervals: the gaps between successive indices may vary and may become arbitrarily large.
Proof. We use induction on \(k\). Since \(n_0\geq0\), the claim holds at \(k=0\). Suppose \(n_k\geq k\). Because \(n_{k+1}\) is an integer strictly larger than \(n_k\), we have \(n_{k+1}\geq n_k+1\). Therefore $$ n_{k+1}\geq n_k+1\geq k+1. $$ This proves \(n_k\geq k\) for every \(k\). Now fix \(M\). Whenever \(k\geq M\), the inequality just proved gives \(n_k\geq k\geq M\), so the subsequence indices eventually pass \(M\). \(\square\)
This fact explains why a subsequence cannot remain confined to a finite initial part of the original sequence. It also lets us use estimates that hold for every sufficiently large original index: eventually, the selected indices fall within the range where those estimates apply.
Checking Convergence Along Selected Indices
To show that a subsequence \((a_{n_k})\) converges to \(L\), we check the epsilon–N definition using \(k\) as the subsequence index. The required threshold is a threshold for \(k\), not necessarily the same threshold used for \(n\). When the original sequence has a useful formula, substitute the selected indices and estimate the resulting error.
Worked Example: A Sparse Subsequence Approaching One
Let \(a_n=n/(n+1)\), and select the indices \(n_k=2k+1\). They are strictly increasing because \(n_{k+1}-n_k=2>0\), so they define a subsequence. Substitution gives $$ a_{n_k}=\frac{2k+1}{2k+2}=1-\frac{1}{2k+2}. $$ We verify that this subsequence converges to \(1\). Given \(\varepsilon>0\), choose \(K\in\mathbb{N}_0\) so large that \(2K+2>1/\varepsilon\). For every \(k\geq K\), we have \(2k+2\geq2K+2>1/\varepsilon\), and hence $$ |a_{n_k}-1|=\frac{1}{2k+2}<\varepsilon. $$ Thus \(a_{n_k}\to1\). The selected terms are spaced two indices apart, but their spacing does not prevent convergence.
The result Subsequences of a Convergent Sequence, established in Convergent Sequences, gives a useful principle in the opposite direction: if the full sequence converges to \(L\), then every subsequence converges to \(L\). The example above verifies convergence of one particular subsequence directly; it does not require the full sequence to have been analyzed first.
Worked Example: Choosing Indices Where a Formula Is Simple
Consider \(a_n=\cos(n\pi/2)\) for \(n\in\mathbb{N}_0\). The full sequence cycles through \(1,0,-1,0\). Choose \(n_k=4k\). Since \(n_k\) is strictly increasing, $$ a_{n_k}=\cos(4k\pi/2)=\cos(2k\pi)=1 $$ for every \(k\). This is a constant subsequence, so it converges to \(1\) by A Constant Sequence Converges.
A different choice, \(m_k=4k+2\), also gives strictly increasing indices, and direct substitution yields $$ a_{m_k}=\cos((4k+2)\pi/2)=\cos((2k+1)\pi)=-1 $$ for every \(k\). This subsequence converges to \(-1\). The two selections show why a sequence can have different convergent subsequences, and why their limits can reveal behavior that prevents the full sequence from converging. The result Distinct Subsequence Limits Obstruct Convergence states this last conclusion.
Repeated Values Produce Constant Subsequences
One especially direct way to find a convergent subsequence is to find a value that occurs at infinitely many indices. Selecting those indices produces a constant subsequence. The argument works whether or not the original sequence takes only finitely many values.
Proof. Let \(E=\{n\in\mathbb{N}_0:a_n=c\}\). By assumption, \(E\) is infinite. Choose its elements in increasing order, writing them as \(n_0<n_1<n_2<\cdots\). Such an increasing selection is possible: after choosing any \(n_k\), the infinite set \(E\) cannot be contained in the finite set \(\{0,1,\ldots,n_k\}\), so there is an element of \(E\) larger than \(n_k\). By the definition of \(E\), \(a_{n_k}=c\) for every \(k\). Thus \((a_{n_k})\) is the constant sequence with value \(c\), and it converges to \(c\) by A Constant Sequence Converges. \(\square\)
Worked Example: Selecting a Repeated Value
Define \(a_n=6\) when \(n\) is divisible by \(5\), and \(a_n=n\) otherwise. For each \(k\in\mathbb{N}_0\), the index \(n_k=5k\) is divisible by \(5\), and these indices are strictly increasing. Therefore $$ a_{n_k}=a_{5k}=6 $$ for every \(k\). This selected sequence converges to \(6\).
The original sequence does not converge to \(6\): at indices \(n=5k+1\), its terms are \(a_{5k+1}=5k+1\), which grow without bound as \(k\) increases. This example emphasizes the distinction between a convergent subsequence and convergence of the full sequence. A subsequence records the behavior along its selected indices only.
Finite-Valued Sequences Always Have a Convergent Subsequence
The repeated-value result gives an immediate guarantee for any sequence that takes values in a finite set. Although the sequence may switch among those values in a complicated order, at least one of them must occur infinitely often. This is a simple but useful subsequence-extraction argument.
Proof. Suppose every term belongs to the finite set \(S=\{c_1,\ldots,c_r\}\). Assume, for contradiction, that each \(c_j\) occurs only finitely many times. Then for each \(j\), the set \(E_j=\{n:a_n=c_j\}\) is finite. Every nonnegative integer \(n\) belongs to one of these sets, since \(a_n\in S\). Consequently, $$ \mathbb{N}_0=E_1\cup E_2\cup\cdots\cup E_r. $$ A finite union of finite sets is finite, so this equality would make \(\mathbb{N}_0\) finite, a contradiction. Hence some value \(c_j\) occurs at infinitely many indices. The theorem An Infinitely Repeated Value Gives a Convergent Subsequence then supplies a subsequence converging to \(c_j\). \(\square\)
The finite-valued hypothesis matters. A sequence whose terms lie in an infinite set need not repeat any value, and the argument above then gives no constant subsequence. The next tutorial, Bolzano-Weierstrass Theorem, develops a much broader guarantee: boundedness, rather than a finite list of possible values, will be enough to obtain a convergent subsequence.
Making a Further Selection
Subsequence selection can be performed more than once. If we first select a subsequence and then select some of its terms, the result is still a subsequence of the original sequence. This is useful when an initial selection reveals a pattern and a second selection isolates a simpler one.
Proof. The indices \(m_0<m_1<m_2<\cdots\) are strictly increasing. Since \(n_0<n_1<n_2<\cdots\), whenever \(m_{j+1}>m_j\) we have \(n_{m_{j+1}}>n_{m_j}\). Thus the indices \(n_{m_0}<n_{m_1}<n_{m_2}<\cdots\) are strictly increasing, which proves that \((a_{n_{m_j}})\) is a subsequence of the original sequence.
If \(a_{n_k}\to L\), apply the earlier theorem Subsequences of a Convergent Sequence to the convergent sequence \((a_{n_k})\) and its subsequence \((a_{n_{m_j}})\). It follows that \(a_{n_{m_j}}\to L\), as claimed. \(\square\)
Worked Example: Selecting Even Positions Twice
Let \(a_n=1/(n+1)\). First take \(n_k=2k\), which gives the subsequence \(a_{2k}=1/(2k+1)\). Now select the even positions within this subsequence by setting \(m_j=2j\). The resulting original indices are $$ n_{m_j}=2(2j)=4j, $$ so the twice-selected sequence is \(a_{4j}=1/(4j+1)\). The indices \(4j\) are strictly increasing, as the composition theorem requires. Since \(a_n\to0\), the theorem Subsequences of a Convergent Sequence shows that both the first selection and the further selection converge to \(0\).
What a Convergent Subsequence Tells Us
A convergent subsequence certifies that the original sequence comes arbitrarily close to its subsequential limit at selected, increasingly late indices. By the definition of cluster point in the previous tutorial, that limit is a cluster point of the original sequence. The Infinite Nearness Criterion for Subsequential Limits gives another way to recognize the same behavior: every neighborhood of the limit contains terms at infinitely many indices.
A convergent subsequence does not by itself show that the full sequence converges. Terms at unselected indices may behave differently, as the repeated-value example demonstrates. Nor must a sequence have only one convergent subsequence limit: the cosine example produced limits \(1\) and \(-1\). When two subsequences converge to distinct real numbers, Distinct Subsequence Limits Obstruct Convergence rules out convergence of the full sequence.
The practical task is therefore to specify the indices carefully, verify that they increase strictly, substitute them into the original terms, and then check the resulting limit. When a value repeats infinitely often, selecting those occurrences is enough. For bounded sequences with no such immediate repetition, a more general construction is needed; that is the purpose of the Bolzano-Weierstrass Theorem.
Check Your Understanding
Use the definition and the proved results to answer the following questions.
- What two properties must an index sequence satisfy in order to define a subsequence?
- Why must the indices \(n_k\) of a subsequence eventually exceed any fixed nonnegative integer \(M\)?
- If a value occurs at infinitely many indices, how can those indices be used to obtain a convergent subsequence?
- Why must some value occur infinitely often when a sequence takes values in a finite set?
- If a convergent subsequence is selected again to form a further subsequence, what happens to its limit?
- Why does the existence of one convergent subsequence not, in general, prove that the full sequence converges?