3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Red Jul 18 '18 at 1:21
  • $\begingroup$ 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. $\endgroup$ – Bernard W Jul 18 '18 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.