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.


1 Answer 1


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 $|z_n - L| < \epsilon$.

  • $\begingroup$ Couldn't get the last part $\endgroup$
    – Soham
    Jan 12, 2020 at 9:50
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 12, 2020 at 10:07

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