# How to prove the sequence $a_0, a_1, \ldots$ converges iff $a_0, a, a_1, a \ldots$ converges?

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.

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How do you expect to reason about infinite sequences without using the definition of infinite sequence? – Chris Eagle Oct 14 '12 at 22:11
+1 for effort. Also, this looks fine to me. For such a general sequence, anything other than applying definitions and work it out from the bottom would probably take significantly more effort. – Arthur Oct 14 '12 at 22:18
@ChrisEagle: Thanks. I feel a little better on my proof now. – Chan Oct 14 '12 at 22:18

Your $\Rightarrow$ proof is erroneous, or at least confused. You can't show that a sequence converges by selecting a convergent subsequence from it; if you could, all sorts of things would converge, such as $\langle 0,1,0,2,0,3,0,4\ldots\rangle$.

Try it like this. We want to show that $\langle a_0, a, a_1, a, a_2, a\ldots\rangle$ converges. Let's call this sequence $\langle b_0, b_1, b_2, \ldots\rangle$. Let $\epsilon>0$ be given. It suffices to show that we can find $N_b$ such that for every $M_b>N_b$, $|b_{M_b} - a|<\epsilon$.

Since $a_i$ converges to $a$, we can find an analogous $N_a$ such that $|a_{M_a} - a|<\epsilon$ for every $n_a>N_a$. Then take $N_b = N_a$. If $n_b > N_b$, then either $b_{n_b}$ is $a$, and so has $|a_{n_a} - a| = 0 <\epsilon$, or $b_{n_b} = a_m$ for some $m > N_b = N_a$, and for that reason we know that $a_m - a| <\epsilon$.

Your $\Leftarrow$ proof is also worrying me, because it seems to me that you gave up and waved your hands at the point where you say "otherwise $|a_n - L| < \epsilon$ would make no sense." That's not a proof, and you can do better. Why will $|a_n - L| < \epsilon$ fail for any $L\ne a$? Can you exhibit a value of $\epsilon$ that makes it fail?

But even then you aren't done. You've shown that $\langle a_0, a, a_1, a, a_2, a\ldots\rangle$ converges to $a$; now you have to show that $\langle a_0, a_1, a_2\ldots\rangle$ converges to $a$. You didn't prove this; you only asserted it. To prove it, you need to give a reason. You claim that for any given $\epsilon$, sufficiently far $a_i$ will have $|a_i - a| < \epsilon$; you gave no indication of how you planned to do this.

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What he does is not selecting a convergent subsequence, he's partitioning into two subsequences and showing that they both converge to the same limit. – Arthur Oct 14 '12 at 22:20
If that's what he thinks he's doing, he should say so! – MJD Oct 14 '12 at 22:21
@MJD: Thanks for the correction. – Chan Oct 14 '12 at 22:29

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.

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The choice $2N + 1$ is very clever. Thank you. – Chan Oct 14 '12 at 22:37
Could I ask you one thing about this statement? "then all subsequences must converge to the same limit", is this a fact? or we have to prove it? Thank you. – Chan Oct 14 '12 at 22:50
Got it. Just want to make sure, I remembered reading in a textbook. – Chan Oct 14 '12 at 22:57