Solution verification of "sequence that converges on even and odds converges"

Ozma

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Oct 14, 2020
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My textbook proves that if a2nαa_{2n} \to \alpha and a2n+1αa_{2n+1} \to \alpha with αR\alpha \in \mathbb{R} then anαa_n \to \alpha considering the max between 2N12N_1 and 2N2+12N_2+1 that make a2nα<ϵ|a_{2n}-\alpha|<\epsilon and a2n+1α<ϵ|a_{2n+1}-\alpha|<\epsilon true for any nmax{2N1,2N2+1}n \ge \max\{2N_1,2N_2+1\} and so it makes true that anα<ϵ|a_n-\alpha|<\epsilon for any nmax{2N1,2N2+1}n \ge \max\{2N_1,2N_2+1\}. Can I prove this by showing that (an)n(a_n)_n is Cauchy, distinguishing for n,mn,m 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 n,mNn,m\in\mathbb{N}. It is anam=anα+αamanα+amα|a_n-a_m|=|a_n-\alpha+\alpha-a_m| \le |a_n-\alpha|+|a_m-\alpha|, now I distinguish three cases. If n,mNn,m\in\mathbb{N} are both even, anα+amα<ϵ|a_n-\alpha|+|a_m-\alpha|<\epsilon if n,mNNn,m \ge N\in\mathbb{N} because by hypothesis limna2n=α\lim_{n \to \infty} a_{2n}=\alpha and so such NNN\in\mathbb{N} exists. Same if n,mn,m are both odd. Considering, without loss of generality, nn even and mm odd it is anamanα+amα=a2kα+a2h+1α|a_n-a_m| \le |a_n-\alpha|+|a_m-\alpha|=|a_{2k}-\alpha|+|a_{2h+1}-\alpha| for some h,kNh,k \in \mathbb{N} and it is a2kα+a2h+1αϵ/2+ϵ/2=ϵ|a_{2k}-\alpha|+|a_{2h+1}-\alpha| \le \epsilon/2+\epsilon/2=\epsilon if h,kmax{n1,n2}h,k \ge \max\{n_1,n_2\} where n1,n2Nn_1,n_2 \in \mathbb{N} are the indexes of the limits hypotheses limna2n=α\lim_{n \to \infty} a_{2n}=\alpha and limna2n+1=α\lim_{n \to \infty} a_{2n+1}=\alpha. So, in every possible case, (an)n(a_n)_n is Cauchy and hence convergent.

Finally, I conclude that the limit of (an)n(a_n)_n is still α\alpha because a convergent sequence must converge to the same limit of its subsequences. Is this reasoning correct?
 
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