So if we have that $a_{2n} \rightarrow a$ and $a_{2n-1} \rightarrow a$, then $a_n \rightarrow a$.
First off, I know that the converse is true by definition since we know if a sequence converges, then all subsequences converge as well.
I think that for this case, since the two subsequences are respectively, the limsup and liminf of the sequences, that I can say that since both converge to a, that the sequence itself will also converge to a but I do not know how to show this in rigorous proofs.
Let $\varepsilon > 0$ be given. We will construct an index $N$ such that $\def\abs#1{\lvert #1 \rvert} \abs{a_n - a} < \varepsilon$ for all $n \geq N$.
Since the even subsequence $a_{2k} \to a$ as $k \to \infty$, there is an index $K_0$ such that $\abs{a_{2k} - a} < \varepsilon$ for all $k \geq K_0$. Similarly, since the odd subsequence $a_{2k-1} \to a$ as $k \to \infty$, there is an index $K_1$ such that $\abs{a_{2k-1} - a} < \varepsilon$ for all $k \geq K_1$.
Now, set $N = \max \{ 2K_0, 2K_1-1 \}$. For every $n \geq N$, if $n$ is even then $n = 2k$ for some $k \geq K_0$, and if $n$ is odd then $n = 2k-1$ for some $k \geq K_1$. Either way, $\abs{a_n - a} < \varepsilon$, as desired.
