I've searched for this but I didn't found anything here, hope it is not a duplicate.
I would like to prove that if $a_n \to l$ for $n \to \infty$ then $a_{n+k} \to l$ for fixed $k\in\mathbb{Z}\setminus \{0\}$.
Let $\varepsilon>0$. By hypothesis $a_n \to l$ for $n \to \infty$, so for all $\varepsilon>0$ exists $N_{\varepsilon} \in \mathbb{N}$ such that if $n>N_{\varepsilon}$ then $|a_n-l|<\varepsilon$. Let $k \in \mathbb{Z}\setminus \{0\}$. Since $n+|k|>n$, when $n>N_{\varepsilon}$ we have that $n+|k|>n>N_{\varepsilon} \Rightarrow n+|k|>N_{\varepsilon}$; so $|a_{n+|k|}-l|<\varepsilon$.
This shows that $a_{n+k} \to l$ for $n \to \infty$ for all fixed $k\in\mathbb{Z}\setminus \{0\}$.
Is this correct? I've used $|k|$ because it unifies both the cases for $k>$ and $k<0$, but I'm not sure if it is correct. Thanks.
The use of $|k|$ is not correct because although it allows you to correctly conclude that $a_{n+|k|}$ converges to $l$, this tells you nothing about $a_{n+k}$, except when $k$ is positive. Conceptually, this argument fails at its core, in that if $n$ is big enough for $a_n$ to be close to $l$, it is not necessarily the case that $a_{n-k}$ is close to $l$, if $k$ is big enough.
If $a_n$ gets close to $l$ starting at $N$, then we need to go to at least $N-k$ in order to make sure $a_{n+k}$ will be close to $l$. For example if $a_n$ gets close to $l$ starting at $N$, then we can only be sure $a_{n-10}$ is that close to $l$ starting at $N+10$. In general, the argument is that $n+k>N$ if and only if $n>N-k$, by simple algebra.