I have to prove that if a sequence ${a_n}$ has the property that both subsequences $a_{2n}$ and $a_{2n+1}$ converge to the same limit $a$, then the original sequence ${a_n}$ also converges to $a$.
My idea was:
We want to show that the sequence ${a_n}$ converges to $a$, i.e., for all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n - a| < \epsilon$ for all $n \geq N$.
Suppose that $a_{2n} \rightarrow a$ and $a_{2n+1} \rightarrow a$ as $n \rightarrow \infty$. Then for all $\epsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that $|a_{2n} - a| < \frac{\epsilon}{2}$ for all $n \geq N_1$ and $|a_{2n+1} - a| < \frac{\epsilon}{2}$ for all $n \geq N_2$.
Let $N := \max{\{2N_1, 2N_2+1\}}$. Then for all $n \geq N$, either $n = 2k$ or $n = 2k+1$ for some $k \in \mathbb{N}$.
For $n = 2k$ with $k \geq N_1$, we have $|a_n - a| = |a_{2k} - a| < \frac{\epsilon}{2}$. For $n = 2k+1$ with $k \geq N_2$, we have $|a_n - a| = |a_{2k+1} - a| < \frac{\epsilon}{2}$. Thus, we have $|a_n - a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ for all $n \geq N$.
Therefore, we have shown that ${a_n}$ converges to $a$, as desired.
Have I done it right?
