Subsequence Convergence Theorem

subsequence

Let $(a_{n})$ be a sequence. Suppose that:

n_{1}, n_{2}, n_{3}, \dots \in N

satisfying:

1 \leq n_{1} < n_{2} < n_{3} < \dots

Then the sequence $(a_{n_{k}})$ for all $k \in N$ is a subsequence of $(a_{n})$ . It's a proper subsequence if $(a_{n_{k}})$ doesn't have all $k = n_{k}$ , namely $\exists k$ where $k \neq n_{k}$ .

The idea is that we just remove possibly some terms, and then preserve the order.

\forall k \in N, n_{k} \geq k

Example

Consider the sequence:

(\frac{1}{n}) = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots)

Some sub sequences of these include:

(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots)

(1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \dots)

(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots)

an invalid sub sequence would be:

(1, \frac{1}{3}, \frac{1}{2}, \frac{1}{4}, \dots)

since the order of terms in the sequence must be preserved.

Notice that each of these sub sequences converges to the same value (namely 0 in this case) to the original sequence. This is what the next theorem tries to show.

The Theorem

Subsequence Convergence Theorem

A subsequence of a convergent sequence converges to the same value as the original sequence.

Proof
Note our observation that $\forall k \in N, n_{k} \geq k$ . Say $(a_{n})$ converges to some $L$ . Then $\forall ε > 0 \exists N \in N$ where for all $n \geq N$ then:

| a_{n} - L | < ε

Let $(a_{n_{k}})$ be a subsequence. Let $ε > 0$ . Then $\exists N \in N$ where for any $n \geq N$ then $| a_{n} - a | < ε$ . Suppose $k \geq N$ . Then by construction then $n_{k} \geq k \geq N$ so then $| a_{n_{k}} - a | < ε$ as desired.
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Example of WCT

Consider the sequence:

(1, - 1, 1, - 1, \dots)

we could use the Convergence of a Sequence definition to show this diverges. But we can also show that the subsequences $a_{2 n}$ and $a_{2 n - 1}$ for all $n \in N$ are valid subsequences. They have different limits of -1 and 1 respectively. If $a_{n}$ itself converged, then these subsequences need to converge to the same limit since Limits are Unique. But they don't, so then by our SCT then the original sequence cannot converge.