# If $\lim_{n \rightarrow \infty}x_n=\lim_{n \rightarrow \infty}y_n=L$ then the sequence $\{x_1,y_1,x_2,y_2,...\}$ tends to $L$

Let $$\lim_{n\rightarrow\infty} x_n = L$$ and $$\lim_{n\rightarrow\infty} y_n = L$$. Then show that the using definition of convergence that sequence $$z_n = (z_1, z_2, z_3, z_4,...) = (x_1, y_1, x_2, y_2, ....)$$ converges to $$L$$.

My attempt:

Clearly, $$|x_n-L|<\epsilon$$ $$\forall n>k_1$$ and $$|y_n-L|<\epsilon$$ $$\forall n>k_2$$.

$$z_{n}$$ can be separated into the cases $$z_{2m}$$ and $$z_{2m+1}$$ where $$m\in\mathbb{N}$$.

We observe, $$z_{2m}=y_m$$ and $$z_{2m+1}=x_{m+1}$$ . Thus, both the odd and even terms of $$z_n$$ converge separately to the same limit $$L$$ since $$x_n$$ and $$y_n$$ converge to $$L$$.

Now, the last part of the logic is not proper and mathematically rigorous. I need help in stating that more formally.

Fix $$\epsilon > 0$$, and suppose for $$n > M_1$$, we have $$|x_n - L| < \epsilon$$ (possible since $$x_n \to L$$). Similarly, suppose for $$n > M_2$$, $$|y_n - L| < \epsilon$$. Choose $$M = 2\max\{M_1, M_2\}$$, and for $$n > M$$ we have $$|z_n - L| < \epsilon$$.
• We have $z_n = x_m$ or $y_m$ for some $m \in \mathbb{Z}^+$. If $z_n = x_m$, then $n = 2m - 1$ so $n > M \implies 2m - 1 > 2M_1 \implies m > M_1$ so $|x_n - L| < \epsilon$. If $z_n = y_m$, then $n = 2m$ so $n > 2M_2 \implies m > M_2$, so a similar conclusion follows. Jan 12, 2020 at 10:07