$a_{2n} \to L$ and $a_{2n-1} \to L$ implies $a_n \to L$

Question

Task: Assume that $a_{2n}$ converges to $L$ and $a_{2n-1}$ converges to $L$, then $a_n$ also converges to $L$.

Proof: Let $a_m$ be an arbitrary subsequence of $a_n$ and $a \in a_m$. Then $a$ must also be an element of either $a_{2n}$ or $a_{2n-1}$ since $a_{2n-1}$ has all the odd and $a_{2n}$ all the even indices. Because both converge to $L$, $a_m$ must too. Thus $a_n \to L$ because every subsequence of $a_n$ converges to $L$.

Does that count as a proof? Thanks for all help in advance!

Answer 1

Because both converge to $L$, $a_m$ must too.

This requires justification, and quite frankly seems to just be using what you want to prove, making your argument somewhat circular. The following should do though:

Let $\varepsilon>0$. As $a_{2n}\to L$ and $a_{2n-1}\to L$ as $n\to\infty$, we can find an $N\in\mathbb{N}$ such that

$$\lvert a_n-L\rvert<\varepsilon$$

if $n=2m$ or $n=2m-1$ for some $m$, and $n\geq N$ (this is just using the definition of the limit, but phrasing it in a bit of a more useful way). But as any $n\in\mathbb{N}$ is on that form (i.e. is either even or odd), it follows that

$$\lvert a_n-L\rvert<\varepsilon$$

for all $n\geq N$.