# Proof the limit of $c_n$ when $c_n=a_n$ for odd n and $c_n=b_n$ for even n

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.

• What can you say about $|c_n-g|$ when $n>\max(N_1,N_2)$? Feb 10, 2020 at 16:18
• 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? Feb 10, 2020 at 16:22
• Exactly.${}{}{}$ Feb 10, 2020 at 16:23
• Thank you! Now it is clear. Feb 10, 2020 at 16:27

## 1 Answer

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 $$n > N_2$$.