Subsequences

A subsequence a_nk consists of selected terms from a sequence a_n, corresponding to certain indices n_k.

A Practical Example

Consider the sequence a_n:

$$ a_n = a_1, a_2, a_3, a_4, \dots $$

Suppose we extract only the terms whose indices are even, that is, where n_k = 2k:

$$ a_{n_k} = a_2, a_4, \dots $$

In this case, a_nk is a subsequence of a_n.

Note. The sequence consisting of the terms with odd indices, where n_k = 2k - 1, is also a subsequence of a_n.

Example 2

Consider the sequence:

$$ a_n = n^2 $$

The first six terms of a_n for k = 1, 2, 3, 4, 5, 6 are:

$$ a_n = 1, 4, 9, 16, 25, 36, \dots $$

Let’s extract the subsequence a_nk defined by the indices n_k = 2k. This means we’re picking out only the terms at even positions.

For example, when k = 1 we have n₁ = 2; when k = 2, we have n₂ = 4, and so on.

The terms of the subsequence extracted from a_n are:

$$ a_{n_k} = 4, 16, 36, \dots $$

Theorems About Subsequences

Theorem 1

If the sequence of natural numbers n_k is strictly increasing, then n_k ≥ k.

In other words, the index n_k of the subsequence a_nk is always at least as large as the index k of the subsequence itself.

Example

Consider a subsequence defined by:

$$ n_k = 2k $$

For k = 1, we have n₁ = 2.

For k = 2, we have n₂ = 4.

And so forth.

Proof

The proof proceeds by mathematical induction. Let’s start with the base case k = 1:

$$ P(1): n_1 \ge 1 $$

In the inductive step, we assume that the following holds:

$$ P(k): n_k \ge k $$

We then need to show that:

$$ P(k+1): n_{k+1} \ge k+1 $$

Since the sequence n_k is strictly increasing, it follows that:

$$ n_{k+1} > n_k $$

Given that n_k ≥ k, we obtain:

$$ n_{k+1} > n_k \ge k $$

Therefore:

$$ n_{k+1} > k $$

Since k is a natural number, its successor is k + 1. Thus, we conclude:

$$ n_{k+1} \ge k+1 $$

Hence, the statement P(k+1) holds true.

This completes the proof by induction.

Theorem 2

If a sequence a_n is strictly increasing and converges to a limit l, then every subsequence extracted from a_n also converges to the same limit l.

Proof

Since the sequence a_n converges, for any ε > 0 there exists an index v such that:

$$ |a_n - l| < \epsilon \quad \forall n > v $$

For any subsequence, we have the property:

$$ n_k \ge k $$

Therefore, whenever k > v, it follows that:

$$ |a_{n_k} - l| < \epsilon $$

And so on.