# $c_n$ is a shuffling of $a_n$ and $b_n$. Prove that $c_n$ converges iff $a_n$ and $b_n$ converge to the same number.

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 $\forall \epsilon \gt 0, \forall n \ge N, |c_n-x| \lt \epsilon$.

• 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). Commented Sep 21, 2016 at 23:50

## 1 Answer

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 $B$.