How to prove $\{a_{n+k}\}^{\infty}_{n=1}$ is convergent and that $\lim\limits_{n\to \infty} a_n=\lim\limits_{n\to \infty} a_{n+k}$?

Question

Let $\{a_{n}\}^{\infty}_{n=1}$ be a sequence of real numbers and let $k$ be a fixed natural number. If $\{a_{n}\}^{\infty}_{n=1}$ is convergent, how to prove $\{a_{n+k}\}^{\infty}_{n=1}$ is convergent and that $\lim\limits_{n\to \infty} a_n=\lim\limits_{n\to \infty} a_{n+k}$.

In my opinion, if $\{a_{n}\}^{\infty}_{n=1}$ is convergent, then $\forall\epsilon\gt 0, \exists N\in \Bbb{Z}^+$ such that if $N\le n$, then $|a_n-L|\lt \epsilon$. I think we can assume the condition to be true and say $\forall\epsilon_1\gt 0, \exists N_1\in \Bbb{Z}^+$ such that if $N_1-k\le n$ (since $k$ is natural number), then $N_1\le n+k$, then $|a_{n+k}-L_1|\lt \epsilon_1$. But I don't see it as a complete proof. Could someone give a clear one?

Answer 1

Your proof is almost complete, if not totally right. However, you can formalize it a bit, as such : If $\{a_{n}\}^{\infty}_{n=1}$ is convergent, then $\forall\epsilon\gt 0, \exists N\in \Bbb{Z}^+$ such that if $N\le n$ then $|a_n-L|\lt \epsilon$. Therefore, the following condition also holds when $N+k\le n+k$. So, for $N_1 = N+k$, we have that if $N_1 \le n+k$, then $|a_{n+k}-L|\lt \epsilon$, for the same $\epsilon$ as before. And therefore, $a_{n+k}$ converges to the same limit as $a_{n}$.

More generally, a similar argument can be made for $a_{f(n)}$ where $f(n)$ is a function of $\Bbb{N} \rightarrow \Bbb{N}$ which is stricly increasing. Just apply the function to the inequality. Hence, every subsequence converges to the same point as the sequence itself, if it is convergent.

Answer 2

Let $k$ be fixed and suppose $a_n \to L$. Consider the sequence $(a_{k+n})_n$. Notice that the convergence of $a_n$ implies that there exists a $N$ such that for all $n \geq N$ we have $|a_n-L|< \epsilon$. Observe then that if $n>N$ then $k+n>N$ and so $|a_{n+k}-L|<\epsilon$ and so we are done. This shows convergence and that they have the same limit $L$.