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Let us consider two sequences $(a_n)$ and $(b_n)$ with equal limits denoted as $g$ for $n \to \infty$. Let us take

$$c_n=\left\{ \begin{array}{ccc} a_n&\mbox{for odd n,}\\ b_n&\mbox{for even n.} \end{array} \right.$$

Using the definition of the limit show that $\lim_{\,n \to \infty} c_n = g$.

Let us take $\epsilon>0.$ Since $\lim_{\,n \to \infty} a_n = g$ and $\lim_{\,n \to \infty} b_n = g$, we have that there is an $N_1 \in \mathbb{N}$ and an $N_2 \in \mathbb{N}$ such that for all $n>N_1$

$$|a_n-g|<\epsilon$$ and for all $n>N_2$ $$|b_n-g|<\epsilon.$$

I do not know how to use this to show that $|c_n-g|$ is also bounded. I would be grateful for any help.

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    $\begingroup$ What can you say about $|c_n-g|$ when $n>\max(N_1,N_2)$? $\endgroup$
    – TonyK
    Feb 10 '20 at 16:18
  • $\begingroup$ It is bounded with $\epsilon$ for odd $n$ (since $|a_n-g|$ is bounded) and it is bounded with $\epsilon$ for even $n$ (since $|b_n-g|$ is bounded), so it is bounded in general. That is it? $\endgroup$
    – Uhans
    Feb 10 '20 at 16:22
  • $\begingroup$ Exactly.${}{}{}$ $\endgroup$
    – TonyK
    Feb 10 '20 at 16:23
  • $\begingroup$ Thank you! Now it is clear. $\endgroup$
    – Uhans
    Feb 10 '20 at 16:27
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Take $N := \max\{N_1,N_2\}$. Then for $n > N$, if $c_n = a_n$ then $|c_n - g| = |a_n - g| < \epsilon$ as $n > N_1$. If $c_n = b_n$ then $|c_n - g| = |b_n - g| < \epsilon$ as $n > N_2$.

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