TITLE: How to prove the sequence $a_0, a_1, \ldots$ converges iff $a_0, a, a_1, a \ldots$ converges?
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: Suppose for every $\epsilon>0$, there is a  positive integer $N$ such that for every $n>N, |a_n-a|<\epsilon$. Let $(b_n)$ be the sequence $(a_0,a,a_1,a,\ldots)$. Then for every $n>2N+1$, it is clear that $|b_n-a|<\epsilon$.
Conversely, if $(b_n)$ converges, then all subsequences must converge to the same limit. Since $(a,a,a,\ldots)$ converges to $a$, the sequence $(a_n)$ does as well.