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Question:

Let ($x_n$) and ($y_n$) be given, and define ($z_n$) to be the "shuffled" sequence
($x_1 ,y_1 ,x_2 ,y_2 ,...,x_n ,y_n ,...$). Prove that ($z_n$) is convergent iff ($x_n$) and ($y_n$) are both convergent with $\lim (x_n) = \lim (y_n)$

My solution:
$\Rightarrow$ Assume $z_n$ is convergent and let $\lim(z_n) =L$. Let $\mathcal{E}>0$ be arbitrary. Because $(z_n) → L$, we can pick $N$ so that $|z_n-L|<\mathcal{E}$ whenever $n \geq N$. Because $y_n= z_{2N}$, then $|y_n-L|<\mathcal{E}$ whenever $n \geq N$. Similarly, because $x_n=z_{2N-1}$, $|x_n-L|<\mathcal{E}$ whenever $n \geq N$. So, both $x_n$ and $y_n$ are convergent with $\lim(x_n) = L = \lim (y_n)$.

$\Leftarrow$ Assume $x_n$ and $y_n$ are convergent, and let $\lim(x_n) = L = \lim(y_n)$. Let $\mathcal{E}>0$ be arbitrary. Choose $N_1$ so that $|x_n-L|< \mathcal{E}$ whenever $n \geq N_1$ and choose $N_2$ so that $|y_n-L|<\mathcal{E}$ whenever $n \geq N_2$. Let $N = \max\{2N_1-1, 2N_2\}$. By construction of $z_n$, $|z_n-L|<\mathcal{E}$ whenever $n \geq N$. So $z_n$ is convergent.

Please let me know if I am on the right track.

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    $\begingroup$ In the converse part, you'd actually need $N = \mathrm{max}(2N_1, 2N_2)$. $\endgroup$ – Daniel Schepler May 10 '17 at 18:55
  • $\begingroup$ Can you explain why? $\endgroup$ – James Snell May 10 '17 at 19:21
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    $\begingroup$ Because $x_n = z_{2n-1}$ and $y_n = z_{2n}$. So actually, $N = \mathrm{max}(2N_1-1, 2N_2)$ would be sufficient. $\endgroup$ – Daniel Schepler May 10 '17 at 19:26
  • $\begingroup$ Oh, I see, thank you! $\endgroup$ – James Snell May 10 '17 at 19:32
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Your proof is fine but there is an easier way

If a sequence is convergent then all subsequences are convergent and converges to the same limit

If $(z_n)$ is convergent then $(x_n)$ and $(y_n)$ are also convergent as they are subsequences of $(z_n)$.

For the converse part, your proof works fine.

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  • $\begingroup$ Ahh, using Bolzano-Weierstass Theorem, right? $\endgroup$ – James Snell May 10 '17 at 19:21

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