Proof verification: another convergent sequence proof Note: Sorry, I posted this earlier with a glaringly obvious error - here's the improved version:
The statement I'm trying to prove is:
Let $ (x_n) $ be a convergent sequence and $ K \in \Bbb N $. Let $ (y_n) $ be the sequence defined by $ y_n = x_{n+K} $. Then $ (y_n) $ is also convergent and we have $ \lim_{n \to \infty}y_n=\lim_{n \to \infty}x_n. $
Proof: 
Let $ \lim_{n \to \infty}x_n=x^* $. Since $(x_n)$ is convergent, by definition we have that given $\epsilon > 0 \quad \exists \quad N \in \Bbb N: \quad \left\lvert x_n - x^* \right\rvert < \epsilon \quad \forall \quad n \ge N$. 
We know that $K \in \Bbb N $, therefore, $ n+K>N $. So we know for definite that
$$ \left\lvert x_{n+K} - x^* \right\rvert < \epsilon \quad \implies \quad \left\lvert y_n - x^* \right\rvert < \epsilon$$
Hence, we can conclude that by definition of convergent sequences, $(y_n)$ is convergent with the limit $x^*$ which is also the limit of $(x_n)$ as $n \to \infty$. So the original statement is true. $\square$
Any confirmation of correctness/corrections would be greatly appreciated. Thank you.
 A: It’s basically correct, but you need to add a little more wording to make your reasoning entirely clear. When you state what it means for the original sequence to converge to $x^*$, you’re not actually picking an arbitrary $\epsilon$ and associated $N$: you’re just saying that for each $\epsilon>0$ there is an $N$ with a certain property. At that point you might continue something like this:

Now let $\epsilon>0$ be arbitrary, and let $N\in\Bbb N$ be such that $|x_n-x^*|<\epsilon$ whenever $n\ge N$. $K\in\Bbb N$, so $K\ge 0$, and therefore $n+K\ge n\ge N$ whenever $n\ge N$. In particular, this ensures that $|x_{n+K}-x^*|<\epsilon$ whenever $n\ge N$. Since $\epsilon>0$ was arbitrary, this implies that $\lim\limits_{n\to\infty}x_{n+K}=x^*$ as well.

A: You work is correct. Notice that every subsequences of a convergence sequence converges to the same limit. By formulates,
If $(x_n)_n$ converges to $a$, then for every  ascending strictly functions $\varphi: \mathbb{N} \rightarrow \mathbb{N}$, the sequences $(x_{\varphi(n)})_n$ is converges to $a$
