Problem
Prove that the sequence $a_0, a_1, a_2, \ldots$ converges to $a$ if and only if the sequence $a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges.
Here is my approach:
$\Rightarrow$:
Since $a_0, a_1, a_2, \ldots$ converges to $a$, by definition of limit,
for every $\epsilon > 0$, $\exists N \in \mathbb{N}$ such that for all
$n > N$, then $|a_n - a| < \epsilon$. Now consider the subsequence
$$a, a, a, a, \ldots$$
We have that $|a - a| < \epsilon, \, \, \forall \epsilon > 0$, thus
$a, a, a, a \ldots$ also converges to $a$. Hence,
$a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges.
$\Leftarrow$:
Suppose that $a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges to $L$,
$L \neq \pm \infty$, by definition of limit, for every $\epsilon > 0$, $\exists N \in \mathbb{N}$
such that for all $n > N$, then $|a_n - a| < \epsilon$, thus there must be a sequence
$$a_{N+1}, a, a_{N+2}, a, a_{N+3}, a, a_{N+4}, \ldots$$
that is getting closer and closer to $L$. But there is always an alternating
$a$ between each $a_i$ and $a_{i+1}$, so $L = a$ otherwise $|a_n - L| < \epsilon$ would
make no sense. Therefore $a_0, a_1, a_2, \ldots$ converges to $a$.
However I still feel it's not complete because all my reasons were based on the definition of infinite sequence. I think there must be a way to give a strong argument for this problem. I wonder if anyone could give me a hint/suggestion on my solution? Thanks.