Difference of sequences tending to infinity tends to infinity 
Let $a_n$ be a sequence such that $a_n \to \infty$ as $n \to \infty$. Show that if $b_k = a_k^2 - 4a_k$, then $b_k \to \infty$ as $k \to \infty$.

As $a_n$ tends to infinity we have that for any $K>0$, there exists $n_0$ such that $n  \ge n_0 \implies a_n> K$.
We want to show that $b_k >K$ also for any $K>0$. Now since $a_n >K$ for any $K$ when $n \ge n_0$ it must be that $a_n > \sqrt{K+4}+2$ for any $K >0$.
Thus $b_k=a_k^2-4a_k > (\sqrt{K+4}+2)^2 - 4(\sqrt{K+4}+2) = K$ whenever $k \ge K$.
Is there some conditions which I should make regarding $k$ here? I have a slight confusion with the indexes $n$ and $k$.
 A: Rewriting with more conventional variable names, $\lim\limits_{n\to\infty}x_n = \infty$ means:
$$\forall \epsilon>0\text{ }(\exists N_{\epsilon}\in \mathbb{N}\text{ }(\forall n>N_{\epsilon}\text{ }(x_n > \epsilon))).\text{ }\text{ }\text{ }\text{ }(\star)$$

We want to show if $\{a_n\}_{n\in\mathbb{N}}$ satisfies $(\star)$, then so does $\{b_n\}_{n\in\mathbb{N}}$, where $b_n:= a_n^2 - 4a_n$.

$\underline{Proof:}$
Let $\epsilon>0$ be given. Since $\{a_n\}_{n\in\mathbb{N}}$ satisfies $(\star)$, we have existence of $N_{\epsilon}$ making $\color{red}{a_n > \epsilon}$ for $n>N_{\epsilon}$.
Shifting the red statement, we have:
$$a_n-4 > \epsilon -4,$$
Then since $a_n>\epsilon>0$, positive scaling gives:
$$a_n(a_n-4) > a_n(\epsilon-4)>\epsilon(\epsilon -4).$$
Rewriting, this says:
$$b_n > \epsilon^2-4\epsilon.$$
Now, define $\color{red}{\epsilon'}$ such that $(\epsilon')^2-4(\epsilon') = \epsilon$. Then by satisfaction of $(\star)$ by $\{a_n\}_{n\in\mathbb{N}}$, we have existence of $N_{\epsilon'}$ yielding $a_n>\epsilon'$, which as we have shown, implies $b_n > \epsilon$. We may use this $N_{\epsilon'}$ to show $\{b_n\}_{n\in\mathbb{N}}$ satisfies $(\star)$. If you like, you can make a distinction by saying $N_{\epsilon,b}:= N_{\epsilon',a}$ to show from which version of $(\star)$ it came from. $\square$
A: More simply by the definition of limit we have that eventually, say for $n>n_0$, $a_n> 5$ then for $k>n_0$ we have
$$b_k = a_k^2 - 4a_k =a_k(a_k-4)>a_k$$
and we can easily conclude.

More rigorously we need to show that
$$\forall M, \; \exists k_0, \; b_k>M,\; \forall k>k_0$$
then set $k_0\ge n_0$ such that $\forall n>n_0$ we have $a_n>|M|+4$, therefore
$$b_k=a_k^2-4a_k=M^2+4|M|>M$$
