My textbook proves that if a2n→α and a2n+1→α with α∈R then an→α considering the max between 2N1 and 2N2+1 that make ∣a2n−α∣<ϵ and ∣a2n+1−α∣<ϵ true for any n≥max{2N1,2N2+1} and so it makes true that ∣an−α∣<ϵ for any n≥max{2N1,2N2+1}. Can I prove this by showing that (an)n is Cauchy, distinguishing for n,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,m∈N. It is ∣an−am∣=∣an−α+α−am∣≤∣an−α∣+∣am−α∣, now I distinguish three cases. If n,m∈N are both even, ∣an−α∣+∣am−α∣<ϵ if n,m≥N∈N because by hypothesis limn→∞a2n=α and so such N∈N exists. Same if n,m are both odd. Considering, without loss of generality, n even and m odd it is ∣an−am∣≤∣an−α∣+∣am−α∣=∣a2k−α∣+∣a2h+1−α∣ for some h,k∈N and it is ∣a2k−α∣+∣a2h+1−α∣≤ϵ/2+ϵ/2=ϵ if h,k≥max{n1,n2} where n1,n2∈N are the indexes of the limits hypotheses limn→∞a2n=α and limn→∞a2n+1=α. So, in every possible case, (an)n is Cauchy and hence convergent.
Finally, I conclude that the limit of (an)n is still α because a convergent sequence must converge to the same limit of its subsequences. Is this reasoning correct?
Specifically, I did this: let n,m∈N. It is ∣an−am∣=∣an−α+α−am∣≤∣an−α∣+∣am−α∣, now I distinguish three cases. If n,m∈N are both even, ∣an−α∣+∣am−α∣<ϵ if n,m≥N∈N because by hypothesis limn→∞a2n=α and so such N∈N exists. Same if n,m are both odd. Considering, without loss of generality, n even and m odd it is ∣an−am∣≤∣an−α∣+∣am−α∣=∣a2k−α∣+∣a2h+1−α∣ for some h,k∈N and it is ∣a2k−α∣+∣a2h+1−α∣≤ϵ/2+ϵ/2=ϵ if h,k≥max{n1,n2} where n1,n2∈N are the indexes of the limits hypotheses limn→∞a2n=α and limn→∞a2n+1=α. So, in every possible case, (an)n is Cauchy and hence convergent.
Finally, I conclude that the limit of (an)n is still α because a convergent sequence must converge to the same limit of its subsequences. Is this reasoning correct?