Definition 2.4.1 (Subsequence). Let \(\{a_n\}_{n=1}^{\infty }\) be a sequence of real numbers. A subsequence of \(\{a_n\}_{n=1}^{\infty }\) is in the form of \(\{a_{n_k}\}_{k=1}^{\infty }\), where \(n_k \in \mathbb {N}\), and \begin {align*} n_1<n_2<n_3< ... \end {align*}
The Following proposition and corollary are important for showing non-existence of the limit for some sequences.
Proposition 5. If \(\{a_n\}_{n=1}^{\infty }\) is a sequence of real numbers such that \(\lim \limits _{n \to \infty }a_n =L\), where \(L \in \mathbb {R}\), then any of its subsequences \(\{a_{n_k}\}_{k=1}^{\infty }\) would also have \begin {align*} \lim _{k\to \infty } a_{n_k} = L \end {align*}
Corollary 14. Given a sequence \(\{a_n\}_{n=1}^{\infty }\) of real numbers, if any of the following holds:
There exist two subsequences \(\{a_{n_k}\}_{k=1}^{\infty }\) and \(\{a_{m_j}\}_{j=1}^{\infty }\) with different limits
There exists a subsequence \(\{a_{n_k}\}_{k=1}^{\infty }\) whose limit does not exist
Then, the limit of \(\{a_n\}_{n=1}^{\infty }\) does not exist.
Now, we prove Proposition 5.
Proof. Because, \(a_n \to L\), and the fact that \(n_k \geq k\), we have \begin {align*} \forall \epsilon >0, \exists N>0 \text { s.t } \forall n >N, |a_n - L| < \epsilon \\ \because n_k \geq k, \forall k > N, n_k > N\\ \implies |a_{n_k} - L| <\epsilon \end {align*}
Hence \(a_{n_k} \to L\). The Corollary is the contrapositive of the proposition, hence directly follows. □