Given: $(a_n)$ and $(b_n)$ are sequences. $c_n=(a_1,b_1,a_2,b_2,...)$

Prove: $c_n$ converges if and only if $a_n$ and $b_n$ converge to the same number.

I have attempted a proof, but I feel like I might be missing steps or explanations. I'd appreciate your feedback.

Part 1:

Let $c_n \to x.$

Then, $\forall \epsilon \gt 0, \exists N \in \mathbb N$ such that $\forall n\ge N, |c_n-x| \lt \epsilon$.

However, since $c_{n \ge N}$ contains $(a_N, b_N, a_{N+1}, b_{N+1},...)$*, this means that $\forall n \ge N, |a_n-x|\lt \epsilon$ and $|b_n-x| \lt \epsilon$.

Thus, $a_n$ and $b_n$ both converge to x.

*I'm not sure that this is a correctly stated sentence. Although I feel it is intuitively correct, I'm not entirely sure how to phrase it in a rigorous manner.

Part 2:

Let $a_n \to x$ and $b_n \to x$.

Then, $\forall \epsilon \gt 0, \exists N_1 \in \mathbb N$ such that $\forall n\ge N_1, |a_n-x| \lt \epsilon$

and $\forall \epsilon \gt 0, \exists N_2 \in \mathbb N$ such that $\forall n\ge N_2, |b_n-x| \lt \epsilon$.

Let $N=max(N_1,N_2)$.

Since $c_{n \ge N}$ contains $(a_N, b_N, a_{N+1}, b_{N+1},...)$, this means that $\forall \epsilon \gt 0, \forall n \ge N, |c_n-x| \lt \epsilon$.

  • $\begingroup$ You are correct with your approach, but I think some of the indices need correction, and the sentence you have flagged needs a little rephrasing. You have got the ideas, but the writing is not as clear as it could be (but it is good, in my opinion). $\endgroup$ Sep 21 '16 at 23:50

I might try proving one of the directions by using its contrapositive.

I think the proof that $a_n \rightarrow L$ and $b_n \rightarrow L$ implies $c_n \rightarrow L$ (hence converges) is reasonably straightforward; as you suggest in your write-up, it follows from considering a max argument.

For the other direction, though, consider the following re-phrasing:

If it is not the case that $a_n$ and $b_n$ tend to the same limit, then $c_n$ does not converge.

In this equivalent form, the proof might be a bit smoother. In particular, if either of $a_n$ or $b_n$ diverges then so will $c_n$; if they converge to $A \not = B$ respectively, then consider trying to get $c_n$ to stay within $|A-B|/2$ of a limit; this will not be possible as the shuffled $a_n$ imply terms arbitrarily close to $A$, while the shuffled $b_n$ imply terms arbitrarily close to $B$.


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