My textbook proves that if [imath]a_{2n} \to \alpha[/imath] and [imath]a_{2n+1} \to \alpha[/imath] with [imath]\alpha \in \mathbb{R}[/imath] then [imath]a_n \to \alpha[/imath] considering the max between [imath]2N_1[/imath] and [imath]2N_2+1[/imath] that make [imath]|a_{2n}-\alpha|<\epsilon[/imath] and [imath]|a_{2n+1}-\alpha|<\epsilon[/imath] true for any [imath]n \ge \max\{2N_1,2N_2+1\}[/imath] and so it makes true that [imath]|a_n-\alpha|<\epsilon[/imath] for any [imath]n \ge \max\{2N_1,2N_2+1\}[/imath]. Can I prove this by showing that [imath](a_n)_n[/imath] is Cauchy, distinguishing for [imath]n,m[/imath] of the Cauchy definition the cases odd-odd, even-even and even-odd? I see that this is more complicated, but I want to ask you if the logic is correct and I'm not doing a wrong proof.
Specifically, I did this: let [imath]n,m\in\mathbb{N}[/imath]. It is [imath]|a_n-a_m|=|a_n-\alpha+\alpha-a_m| \le |a_n-\alpha|+|a_m-\alpha|[/imath], now I distinguish three cases. If [imath]n,m\in\mathbb{N}[/imath] are both even, [imath]|a_n-\alpha|+|a_m-\alpha|<\epsilon[/imath] if [imath]n,m \ge N\in\mathbb{N}[/imath] because by hypothesis [imath]\lim_{n \to \infty} a_{2n}=\alpha[/imath] and so such [imath]N\in\mathbb{N}[/imath] exists. Same if [imath]n,m[/imath] are both odd. Considering, without loss of generality, [imath]n [/imath] even and [imath]m[/imath] odd it is [imath]|a_n-a_m| \le |a_n-\alpha|+|a_m-\alpha|=|a_{2k}-\alpha|+|a_{2h+1}-\alpha|[/imath] for some [imath]h,k \in \mathbb{N}[/imath] and it is [imath]|a_{2k}-\alpha|+|a_{2h+1}-\alpha| \le \epsilon/2+\epsilon/2=\epsilon[/imath] if [imath]h,k \ge \max\{n_1,n_2\} [/imath] where [imath]n_1,n_2 \in \mathbb{N}[/imath] are the indexes of the limits hypotheses [imath]\lim_{n \to \infty} a_{2n}=\alpha[/imath] and [imath]\lim_{n \to \infty} a_{2n+1}=\alpha[/imath]. So, in every possible case, [imath](a_n)_n[/imath] is Cauchy and hence convergent.
Finally, I conclude that the limit of [imath](a_n)_n[/imath] is still [imath]\alpha[/imath] because a convergent sequence must converge to the same limit of its subsequences. Is this reasoning correct?
Specifically, I did this: let [imath]n,m\in\mathbb{N}[/imath]. It is [imath]|a_n-a_m|=|a_n-\alpha+\alpha-a_m| \le |a_n-\alpha|+|a_m-\alpha|[/imath], now I distinguish three cases. If [imath]n,m\in\mathbb{N}[/imath] are both even, [imath]|a_n-\alpha|+|a_m-\alpha|<\epsilon[/imath] if [imath]n,m \ge N\in\mathbb{N}[/imath] because by hypothesis [imath]\lim_{n \to \infty} a_{2n}=\alpha[/imath] and so such [imath]N\in\mathbb{N}[/imath] exists. Same if [imath]n,m[/imath] are both odd. Considering, without loss of generality, [imath]n [/imath] even and [imath]m[/imath] odd it is [imath]|a_n-a_m| \le |a_n-\alpha|+|a_m-\alpha|=|a_{2k}-\alpha|+|a_{2h+1}-\alpha|[/imath] for some [imath]h,k \in \mathbb{N}[/imath] and it is [imath]|a_{2k}-\alpha|+|a_{2h+1}-\alpha| \le \epsilon/2+\epsilon/2=\epsilon[/imath] if [imath]h,k \ge \max\{n_1,n_2\} [/imath] where [imath]n_1,n_2 \in \mathbb{N}[/imath] are the indexes of the limits hypotheses [imath]\lim_{n \to \infty} a_{2n}=\alpha[/imath] and [imath]\lim_{n \to \infty} a_{2n+1}=\alpha[/imath]. So, in every possible case, [imath](a_n)_n[/imath] is Cauchy and hence convergent.
Finally, I conclude that the limit of [imath](a_n)_n[/imath] is still [imath]\alpha[/imath] because a convergent sequence must converge to the same limit of its subsequences. Is this reasoning correct?
