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Given sequences $(x_n)$ and $(y_n)$, define $(z_n)$ as $z_{2n-1} = x_n$ and $z_{2n} = y_n$. If $\lim x_n = \lim y_n = a$, so $\lim z_n = a$.

I would like to know if my attempt and writing is correct, thanks in advance!

My attempt

$\lim x_n = a$, in other words, given $\varepsilon_1 > 0$, exists a $N_1 \in \mathbb{N}$ such that $|x_n - a| < \varepsilon_1, \forall n>N_1$.

$\lim y_n = a$, in other words, given $\varepsilon_2 > 0$, exists a $N_2\in\mathbb{N}$ such that $|y_n - a|<\varepsilon_2, \forall n>N_2$.

$|x_n - a| < \varepsilon_1, \forall n>N_1 \implies |z_{2n-1} - a| < \varepsilon_1, \forall n>N_1$.

$|y_n - a|<\varepsilon_2, \forall n>N_2\implies |z_{2n} - a| < \varepsilon_2, \forall n>N_2$.

Define $\epsilon_3=\min\{\varepsilon_1,\varepsilon_2\}$ and $N_3=\max\{N_1, N_2\}$, therefore, $|z_n - a|<\varepsilon_3,\forall n>N_3 \implies |z_n - a|<\varepsilon_3, \forall n>N_3$, in other words, $\lim z_n=a$.

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It's almost correct. You have to show that given $\varepsilon > 0$ you can find an $N$ such that for all $n\geq N$, $|z_n-a|<\varepsilon$. So choose $\varepsilon >0$. Take $N_1, N_2$ such that for all $n\geq N_1$ and all $m\geq N_2$, $|x_n-a|<\varepsilon$ and $|y_n-a|<\varepsilon$. Take $N=4\max\left\{N_1,N_2\right\}$, and let $n\geq N$. Then either $n=2k-1$ or $n=2k$ for some $k\in \mathbb{N}$ with $k\geq N_1$ and $k\geq N_2$. Hence $|z_n-a|=|x_k-a|<\varepsilon$ if $n=2k-1$ and $|z_n-a|=|y_k-a|<\varepsilon$ if $n=2k$.

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  • $\begingroup$ I think I understood your answer, my mistake is that I'm taking $n=k$, but I can't do this because $z_{2n-1} = x_n$ and $z_{2n} = x_n$ are defined on this form and about the $\epsilon$ I can take the same $\epsilon$ instead of $\epsilon_1$ and $\epsilon_2$ because if $\epsilon_2 > \epsilon_1$ for a $N_2 \in \mathbb{N}$, so exists a $N'_2 \in \mathbb{N}$ that $|y_n - a| < \epsilon_1$ for $n > N'_2$ is for this reason that can I take the same $\epsilon$ for the sequences $(x_n)$ and $(y_n)$? On this $N$, I didn't understand why you take $N = 4\max \{N_1,N_2\}$. $\endgroup$ – George Apr 8 '16 at 14:34
  • $\begingroup$ Could you edit your comment? It's hard to read like this. Anyway I take $N=4\max\left\{N_1,N_2\right\}$ to be sure that $N$ is large enough. The point is that you need that for $n\geq N$ and $n=2k-1$ or $n=2k$, you want that $k$ is larger than both $N_1$ and $N_2$. $\endgroup$ – Mathematician 42 Apr 9 '16 at 6:15
  • $\begingroup$ I think I understood your answer, my mistake is that I'm taking $n=k$, but I can't do this because $z_{2n−1} = x_n$ and $z_{2n} = y_n$ are defined on this form and about the $\epsilon$ I can take the same $\epsilon$ instead of $\epsilon_1$ and $\epsilon_2$ because if $\epsilon_2$ > $\epsilon_1$ for a $\mathbb{N_2} \in \mathbb{N}$, so exists a $\mathbb{N'_2} \in \mathbb{N}$ that $|yn−a|$ < $\epsilon_1$ for $n$ > $\mathbb{N'_2}$ $\endgroup$ – George Apr 9 '16 at 23:21
  • $\begingroup$ Is for this reason that can I take the same $\epsilon$ for the sequences $(x_n)$ and $(y_n)$? On this N, I didn′t understand why you take N = 4max.{$N_1,N_2$}. P.S.: I couldn't edit the first comment, I don't know why. $\endgroup$ – George Apr 9 '16 at 23:26

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