Subsequences
A subsequence \( (a_{n_k}) \) of a sequence \( (a_n) \) is obtained by selecting terms whose indices \( (n_k) \) form a strictly increasing sequence of natural numbers.
In simple terms, a subsequence is created by picking some elements from a sequence while keeping their original order.
Starting from a sequence
\[ (a_n) \]
we choose indices such that
\[ n_1 < n_2 < n_3 < \dots \]
This produces a new sequence
\[ (a_{n_1}, a_{n_2}, a_{n_3}, \dots) \]
which is called a subsequence of \( (a_n) \).
A first example
Consider the sequence \( (a_n) \)
$$ a_n = a_1, a_2, a_3, a_4, \dots $$
If we keep only the terms with even indices, defined by \( n_k = 2k \), we obtain
$$ a_{n_k} = a_2, a_4, \dots $$
This is a subsequence of the original sequence.
Note. If instead we take the odd indices \( n_k = 2k - 1 \), we obtain another valid subsequence.
Example 2
Consider the sequence
$$ a_n = n^2 $$
The first terms are
$$ a_n = 1, 4, 9, 16, 25, 36, \dots $$
If we select only even indices, \( n_k = 2k \), we get
For example, \( n_1 = 2 \), \( n_2 = 4 \), and so on.
The subsequence becomes
$$ a_{n_k} = 4, 16, 36, \dots $$
Subsequence limit theorem
If a sequence \( (a_n) \) converges to a finite limit \( l \in \mathbb{R} \), or diverges to \( +\infty \) or \( -\infty \), then every subsequence behaves in the same way, converging or diverging to the same limit.
This means that subsequences cannot "escape" the limiting behavior of the original sequence.
If \( a_n \to l \), then every subsequence \( (a_{n_k}) \) also satisfies \( a_{n_k} \to l \).
If instead \( a_n \to +\infty \) or \( a_n \to -\infty \), all subsequences diverge in the same direction.
Note. This result applies only when the sequence is regular, meaning it either converges or diverges. If the sequence is irregular, its subsequences can behave differently. For example, the sequence \( a_n = (-1)^n \) oscillates: $$ (-1)^n = -1, +1, -1, +1, \dots $$ It does not converge. However, the subsequence with even indices is constant: $$ (-1)^{2n} = +1, +1, +1, \dots $$ and therefore converges. This shows that a convergent subsequence does not imply that the original sequence is convergent.
Let’s look at some examples to better understand these ideas.
Example 1
Consider the sequence
\[ a_n = \frac{1}{n} \]
The terms are
\[ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \]
The limit is
\[ \lim_{n \to \infty} a_n = \frac{1}{n} \to 0 \]
If we take the subsequence with even indices, we get
\[ \lim_{n \to \infty} a_{2n} = \frac{1}{2n} \to 0 \]
If we take square indices, we get
\[ \lim_{n \to \infty} a_{n^2} = \frac{1}{n^2} \to 0 \]
All subsequences converge to the same limit.
Example 2
Consider the sequence
\[ a_n = n + 1 \]
The terms are
\[ 2,3,4,5,6,\dots \]
This sequence diverges to \( +\infty \)
\[ \lim_{n \to \infty} a_n = n+1 \to +\infty \]
The subsequence with even indices behaves in the same way
\[ \lim_{n \to \infty} a_{2n} = 2n + 1 \to +\infty \]
The same is true for square indices
\[ \lim_{n \to \infty} a_{n^2} = n^2 + 1 \to +\infty \]
All subsequences diverge to \( +\infty \).
Example 3
Consider the sequence
\[ a_n = -n \]
The terms are
\[ -1,-2,-3,-4,\dots \]
This sequence diverges to \( -\infty \)
\[ \lim_{n \to \infty} a_n = -n \to -\infty \]
The subsequences behave in the same way
\[ \lim_{n \to \infty} a_{2n} = -2n \to -\infty \]
All subsequences diverge to \( -\infty \).
Example 4
Consider the sequence
\[ a_n = (-1)^n \]
This sequence oscillates and does not have a limit
\[ -1,1,-1,1,-1,1,\dots \]
If we take the subsequence with even indices
\[ \lim_{n \to \infty} a_{2n} = (-1)^{2n} \to +1 \]
If we take the subsequence with odd indices
\[ \lim_{n \to \infty} a_{2n+1} = (-1)^{2n+1} \to -1 \]
In this case, different subsequences converge to different limits.
Further results on subsequences
Theorem 1
If \( (n_k) \) is a strictly increasing sequence of natural numbers, then \( n_k \ge k \) for every \( k \in \mathbb{N} \).
This means that the indices of a subsequence cannot grow more slowly than the natural numbers.
Example
If \( n_k = 2k \), then
for \( k=1 \), \( n_1=2 \);
for \( k=2 \), \( n_2=4 \);
and so on.
Proof
The proof is by mathematical induction.
Base case:
$$ P(1): n_1 \ge 1 $$
Inductive hypothesis:
$$ P(k): n_k \ge k $$
Inductive step:
Since the sequence \( (n_k) \) is strictly increasing,
$$ n_{k+1} > n_k $$
Using the inductive hypothesis,
$$ n_{k+1} > n_k \ge k $$
Therefore,
$$ n_{k+1} \ge k+1 $$
This completes the proof.
Theorem 2
If a sequence \( (a_n) \) is strictly increasing and converges to a limit \( l \), then every subsequence \( (a_{n_k}) \) also converges to \( l \).
Proof
Since \( (a_n) \) converges, for every \( \epsilon > 0 \) there exists \( N \in \mathbb{N} \) such that
$$ |a_n - l| < \epsilon \quad \forall n > N $$
Because \( n_k \ge k \), if \( k > N \) then also \( n_k > N \).
Therefore,
$$ |a_{n_k} - l| < \epsilon $$
Hence, \( a_{n_k} \to l \).
And so on.
