Subsequences of a convergent sequence converge to the same limit as the original sequence.


For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Proof:

Let \((a_n)\) be a sequence and \((a_{n_k})\) be a subsequence. We know that \((a_n)\) convergences to some number \(l\). By definition (2.2.3), we know that there exists an \(N \in \mathbf{N}\) such that if \(n \geq N\), then for any \(\epsilon > 0\), we have

$$ \begin{align*} |a_n - l| \leq \epsilon \quad \text{whenever $n \geq N$}. \end{align*} $$

To prove that \((a_{n_k})\) also converges to the same limit, Choose \(k > N\). Since \(n_k \geq k\) for all \(k\). (for why see, this), then \(n_k \geq N\). This means that for all \(\epsilon > 0\), we must have

$$ \begin{align*} |a_{n_k} - l| \leq \epsilon \quad \text{whenever $k \geq N$}. \end{align*} $$

As we wanted to show. \(\blacksquare\)

References: