Multiplying convergent sequence by divergent sequences of natural numbers

Question

Let $\{x_k\}$ be a sequence converging to the limit $a \ne 0$ and let $\{n_k\} \subset \mathbb{N} $ be a divergent sequence of natural numbers. I want to show that the sequence $\{x_k n_k\}$ is divergent. It seemed easy but I could not prove it.

Answer 1

Proof for the case that $n_k$ is bounded:

$n_k$ is divergent and natural, so for each $N \in \mathbb{N}$ there are $p>q>N$ s.t $|n_p-n_q| \ge 1$

$n_k$ is bounded so let $M \ge 1$ be an upper bound

Let $\epsilon = \frac{|a|}{4M}$ Since $x_n \rightarrow a$, there is a $K$ s.t $|x_n-a|<\epsilon$ for all $n>K$. Also there are $p,q>K$ s.t. $|n_p-n_q| \ge 1$, This implies:

$$|n_px_p-n_qx_q|>|a(n_p-n_q)|-\epsilon(n_p+n_q)> |a| - 2M\epsilon = |a| - \frac{|a|}{2M} $$

And therefore $\{x_kn_k\}$ is divergent.