On this page we shall study subsequences, that is, sequences obtained by extracting some of the terms of a given sequence without altering the order in which they occur.
The notion of subsequence is central to the study of numerical sequences, since it allows one to analyse the behaviour of a sequence by observing only a portion of its terms. Subsequences are especially useful when investigating the convergence, divergence and oscillation of a sequence.
Throughout the article we shall consider real sequences, that is, sequences of the form
\[ a:\mathbb{N}\to\mathbb{R}, \]
and we shall denote their terms by \(a_n\), as \(n\) ranges over \(\mathbb{N}\).
Throughout the text we assume that \(\mathbb{N}=\{0,1,2,\dots\}\).
Contents
- Definition of a subsequence
- Intuitive interpretation
- Examples of subsequences
- How to recognise a subsequence
- Every subsequence of a convergent sequence converges to the same limit
- Subsequences and non-convergence
- Subsequences of sequences diverging to infinity
Definition of a subsequence
Let \((a_n)\) be a real sequence. A subsequence of \((a_n)\) is a sequence obtained by selecting a strictly increasing sequence of natural-number indices
\[ k_0<k_1<k_2<\dots \]
and taking the corresponding terms of the original sequence:
\[ a_{k_0},a_{k_1},a_{k_2},\dots \]
In symbols, a subsequence of \((a_n)\) is often written as
\[ (a_{k_n}), \]
where \((k_n)\) is a sequence of natural numbers such that
\[ k_n<k_{n+1} \]
for every \(n\in\mathbb{N}\).
The condition \(k_n<k_{n+1}\) is essential: it means that the chosen indices must increase strictly. In this way the terms are extracted from the original sequence while preserving their natural order.
An equivalent definition
A subsequence of \((a_n)\) may equally well be defined by means of a strictly increasing function
\[ \varphi:\mathbb{N}\to\mathbb{N}. \]
In this case the subsequence is
\[ (a_{\varphi(n)}). \]
The two definitions are equivalent: it suffices to set
\[ k_n=\varphi(n). \]
In either case, the crucial point is that the indices must increase strictly.
Warning. A subsequence is not a sequence formed by picking terms at random. The selected terms must respect the order in which they appear in the original sequence.
For instance, if we consider the terms
\[ a_5,a_2,a_8, \]
they cannot form the beginning of a subsequence, because the indices are not in increasing order.
The terms
\[ a_2,a_5,a_8, \]
on the other hand, can form the beginning of a subsequence, because the indices \(2,5,8\) are strictly increasing.
Intuitive interpretation
A subsequence is obtained by selecting terms of the original sequence and keeping them in the same order.
For example, given a sequence
\[ a_0,a_1,a_2,a_3,a_4,a_5,\dots, \]
we may select only the terms with even index:
\[ a_0,a_2,a_4,a_6,\dots \]
This is a subsequence of the original sequence.
We may equally select only the terms with odd index:
\[ a_1,a_3,a_5,a_7,\dots \]
This too is a subsequence.
In general, a subsequence observes the original sequence along an increasing sequence of indices.
A property of the indices of a subsequence
If \((k_n)\) is a strictly increasing sequence of natural numbers, then
\[ k_n\ge n \]
for every \(n\in\mathbb{N}\).
Proof. Since \(k_0\in\mathbb{N}\), we have
\[ k_0\ge 0. \]
Moreover, since \((k_n)\) is strictly increasing and takes natural-number values, the inequality \(k_n<k_{n+1}\) yields
\[ k_{n+1}\ge k_n+1. \]
We prove by induction that \(k_n\ge n\) for every \(n\in\mathbb{N}\).
For \(n=0\) we have already observed that \(k_0\ge 0\). Suppose now that \(k_n\ge n\). Then
\[ k_{n+1}\ge k_n+1\ge n+1. \]
Hence, by the principle of induction,
\[ k_n\ge n \]
for every \(n\in\mathbb{N}\).
This property shows that the indices of a subsequence necessarily tend to infinity.
Examples of subsequences
Let us look at a few basic examples.
Example 1. Consider the sequence
\[ a_n=n. \]
If we select the even indices, namely
\[ k_n=2n, \]
we obtain the subsequence
\[ a_{k_n}=a_{2n}=2n. \]
Thus
\[ (a_{2n})=(0,2,4,6,\dots). \]
This is a subsequence of \((a_n)\), because the indices
\[ 0,2,4,6,\dots \]
are strictly increasing.
Example 2. Consider the sequence
\[ a_n=(-1)^n. \]
Selecting the even indices \(k_n=2n\), we obtain
\[ a_{k_n}=a_{2n}=(-1)^{2n}=1. \]
Hence the subsequence of even-indexed terms is
\[ (a_{2n})=(1,1,1,1,\dots). \]
If instead we select the odd indices \(k_n=2n+1\), we obtain
\[ a_{k_n}=a_{2n+1}=(-1)^{2n+1}=-1. \]
Hence the subsequence of odd-indexed terms is
\[ (a_{2n+1})=(-1,-1,-1,-1,\dots). \]
This example is particularly instructive: the sequence \(((-1)^n)\) does not converge, yet it possesses convergent subsequences.
Example 3. Consider the sequence
\[ a_n=\frac{1}{n+1}. \]
If we select the indices
\[ k_n=n^2, \]
we obtain
\[ a_{k_n}=a_{n^2}=\frac{1}{n^2+1}. \]
Hence \(\left(\displaystyle\frac{1}{n^2+1}\right)\) is a subsequence of \(\left(\displaystyle\frac{1}{n+1}\right)\).
The chosen indices are
\[ 0,1,4,9,16,\dots \]
and they form a strictly increasing sequence. Indeed,
\[ 0<1<4<9<16<\dots. \]
Hence the sequence \((a_{n^2})\) is indeed a subsequence of \((a_n)\).
How to recognise a subsequence
To determine whether a sequence \((b_n)\) is a subsequence of a sequence \((a_n)\), one must check whether there exists a strictly increasing sequence of natural-number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
Example. Consider the sequence
\[ a_n=n^2. \]
The sequence
\[ b_n=(n+1)^2 \]
is a subsequence of \((a_n)\). Indeed, it suffices to take
\[ k_n=n+1. \]
Then
\[ a_{k_n}=a_{n+1}=(n+1)^2=b_n. \]
Since \(k_n=n+1\) is strictly increasing, \((b_n)\) is a subsequence of \((a_n)\).
Counterexample. Consider again the sequence
\[ a_n=n. \]
The sequence
\[ b_n=-n \]
is not a subsequence of \((a_n)\), because every term of \((a_n)\) is a natural number, whereas \(b_n\) takes negative values for \(n\ge 1\).
Consequently, there can be no sequence of natural-number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
Warning. It is not enough for the terms of \((b_n)\) to belong to the set of values attained by \((a_n)\). They must also appear in the correct order and be associated with a strictly increasing sequence of indices.
For example, consider
\[ a_n=(-1)^n. \]
The constant sequence \(b_n=1\) is a subsequence of \((a_n)\), since it is obtained by selecting the even indices.
The sequence
\[ 1,-1,1,-1,\dots, \]
on the other hand, is the sequence itself, that is, the trivial subsequence obtained by taking \(k_n=n\).
Every subsequence of a convergent sequence converges to the same limit
The most important result concerning subsequences is the following.
Theorem. If a real sequence \((a_n)\) converges to a real number \(\ell\), then every subsequence \((a_{k_n})\) of it converges to the same limit \(\ell\).
Proof. Suppose that
\[ a_n\to \ell. \]
Let \((a_{k_n})\) be a subsequence of \((a_n)\), with \((k_n)\) strictly increasing.
By the definition of limit, for every \(\varepsilon>0\) there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |a_n-\ell|<\varepsilon. \]
Since \((k_n)\) is a strictly increasing sequence of natural numbers, we have
\[ k_n\ge n \]
for every \(n\in\mathbb{N}\).
Hence, if \(n\ge N\), then
\[ k_n\ge n\ge N. \]
Applying the definition of limit to the index \(k_n\), we obtain
\[ |a_{k_n}-\ell|<\varepsilon. \]
This holds for every \(n\ge N\). Therefore
\[ a_{k_n}\to \ell. \]
We have thus shown that every subsequence of a convergent sequence converges to the same limit as the original sequence.
A consequence
If a sequence possesses two subsequences converging to different limits, then the original sequence does not converge.
Proof. Suppose, for contradiction, that \((a_n)\) converges to a real number \(\ell\).
Then, by the theorem just proved, every subsequence of \((a_n)\) would have to converge to \(\ell\).
But if there exist two subsequences converging to two different limits, we reach a contradiction.
Hence the sequence \((a_n)\) cannot converge.
Subsequences and non-convergence
Subsequences provide a very effective method for proving that a sequence does not converge.
The idea is simple: if we can find two subsequences of the same sequence that converge to different limits, then the original sequence cannot have a limit.
A fundamental example
Consider the sequence
\[ a_n=(-1)^n. \]
The subsequence of even indices is
\[ a_{2n}=(-1)^{2n}=1. \]
Hence
\[ a_{2n}\to 1. \]
The subsequence of odd indices is
\[ a_{2n+1}=(-1)^{2n+1}=-1. \]
Hence
\[ a_{2n+1}\to -1. \]
We have found two subsequences of the same sequence that converge to two different limits:
\[ 1\ne -1. \]
Therefore the sequence \(((-1)^n)\) does not converge.
A further example
Consider the sequence
\[ a_n=\frac{1+(-1)^n}{2}. \]
For the even indices we have
\[ a_{2n}=\frac{1+(-1)^{2n}}{2}=\frac{1+1}{2}=1. \]
For the odd indices we have
\[ a_{2n+1}=\frac{1+(-1)^{2n+1}}{2}=\frac{1-1}{2}=0. \]
Hence
\[ a_{2n}\to 1 \]
while
\[ a_{2n+1}\to 0. \]
Since the two subsequences converge to different limits, the sequence \((a_n)\) does not converge.
Warning. Finding one convergent subsequence is not enough to conclude that the original sequence converges.
For instance, the sequence \(((-1)^n)\) possesses the constant subsequence
\[ a_{2n}=1, \]
which converges to \(1\). Nevertheless, the sequence \(((-1)^n)\) does not converge.
To establish the convergence of the original sequence it is not enough to examine a single subsequence: one must control the behaviour of all the terms, possibly by means of suitable criteria.
Subsequences of sequences diverging to infinity
For sequences diverging to infinity there is likewise an important connection with subsequences.
Theorem. If \(a_n\to+\infty\), then every subsequence \((a_{k_n})\) diverges to \(+\infty\).
Proof. Suppose that \(a_n\to+\infty\).
By definition, for every \(M\in\mathbb{R}\) there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ a_n>M. \]
Let \((a_{k_n})\) be a subsequence of \((a_n)\). Since \(k_n\ge n\), if \(n\ge N\) then
\[ k_n\ge n\ge N. \]
Hence
\[ a_{k_n}>M. \]
This holds for every \(n\ge N\). Therefore
\[ a_{k_n}\to +\infty. \]
The case \(a_n\to -\infty\)
In an analogous way, if
\[ a_n\to -\infty, \]
then every subsequence \((a_{k_n})\) diverges to \(-\infty\).
Indeed, for every \(m\in\mathbb{R}\), the hypothesis \(a_n\to -\infty\) ensures that there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ a_n<m. \]
If \(n\ge N\), then \(k_n\ge n\ge N\), and hence
\[ a_{k_n}<m. \]
Therefore
\[ a_{k_n}\to -\infty. \]
Conclusion
Subsequences are a fundamental tool for studying the asymptotic behaviour of a sequence, since they allow us to isolate significant portions of the sequence without altering the order of its terms.
To summarise, a subsequence of \((a_n)\) is a sequence of the form
\[ (a_{k_n}), \]
where \((k_n)\) is a strictly increasing sequence of natural numbers.
If a sequence converges to a real limit \(\ell\), then every subsequence of it converges to the same limit \(\ell\). Consequently, if a sequence possesses two subsequences converging to different limits, then the sequence does not converge.
Conversely, the existence of a convergent subsequence does not imply that the original sequence converges, as the example of the sequence \(((-1)^n)\) shows.
Subsequences are therefore fundamental both for recognising the local behaviour of a sequence and for proving rigorously that it fails to converge.