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.