# Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the “shuffled” sequence…

Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the "shuffled" sequence $(x_1,y_1,x_2,y_2,x_3,y_3,..., x_n,y_n,...)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent and $\lim x_n=\lim y_n$.

Can someone prove this theorem for me? I am lost and all I have so far is: So we can write that $(z_{2n-1})=(x_n)$ and $(z_{2n})=(y_n)$ for $n \in \mathbb N.$

$(\Longrightarrow)$ (Is this correct?)

Assume for $n \ge N_1$ $\Rightarrow$ $|x_{n}-l|< \epsilon$ and assume for $n \ge N_2$ $\Rightarrow$ $|y_{n}-l|< \epsilon$. Choose $M \ge max\{N_1, N_2\}$ and then for $n \ge M$ for $2n$ and $2n+1$ is we get $|z_n -l| \lt \epsilon$.

($\Longleftarrow$) I don't have anything for this direction.

Note that this question If $x_n \to a$ and $x'_n \to a$, then $\{x_1, x'_1, x_2, x'_2, ...\} \to a$ only addresses the theorem in one direction so it isn't exactly a duplicate.

Let $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ be two sequences of real numbers. We define the sequence $\{z_n\}_{n=1}^\infty$ by setting $z_{2n-1}:=x_n$ and $z_{2n}:=y_n$ for each $n=1,2,\ldots$.

$(\Longleftarrow)$ Suppose $\underset{n \to \infty}{\lim}z_n=L.$ Let $\varepsilon>0$ be given. Since $z_n \to L$ as $n \to \infty$, there is a positive integer $N$ so that \begin{equation} |z_n-L|<\varepsilon \text{ whenever } n \geq N. \end{equation} Now notice that if $n \geq N$, then $2n-1 \geq N$ and $2n \geq N$. Thus we have \begin{aligned} &|x_n-L|=|z_{2n-1}-L|<\varepsilon \text{, and } \\& |y_n-L|=|z_{2n}-L|<\varepsilon \text{ whenever } n \geq N. \end{aligned} Therefore, $\underset{n \to \infty}{\lim}x_n=L=\underset{n \to \infty}{\lim}y_n$.

$(\Longrightarrow)$ Suppose $\underset{n \to \infty}{\lim}x_n=L=\underset{n \to \infty}{\lim}y_n$. Let $\varepsilon>0$ be given. Since the sequences $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that \begin{equation} |x_n-L|<\varepsilon \text{ whenever } n \geq N_1 \end{equation} and \begin{equation} |y_n-L|<\varepsilon \text{ whenever } n \geq N_2. \end{equation} Set $N=\max\{2N_1, 2N_2\}$. Thus we have \begin{equation} |z_n-L|<\varepsilon \text{ whenever } n \geq N. \end{equation} Therefore, $\{z_n\}_{n=1}^\infty$ is a convergent sequence, by definition.

Sometimes it helps to prove a more general statement.

Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.

In this case both ${x_n}$ and ${y_n}$ are subsequences of $z_n$ so the proof will be done.

• I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly. – Red Jul 18 '18 at 1:21
• What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice. – Bernard W Jul 18 '18 at 1:24