If $(x_n)$ is a bounded infinite sequence, prove that $\lim\limits_{n→∞}\frac{x_n}{n^k}=0$

Suppose that $$(x_n)$$ is a bounded infinite sequence of real numbers. Prove that for every $$k \in \mathbb N$$, $$\lim_{n\to \infty} \frac{x_n}{n^k} = 0.$$ You may use the fact that $$\lim\limits_{n→∞}\dfrac{1}{n^k}=0$$ for every $$k \in \mathbb N$$, but do not use any other limit laws.

My attempt:
Let $$ε>0$$ be given. Then $$\left|\dfrac{x_n}{n^k}-0\right|=|x_n|\dfrac{1}{n^k}$$. Now since $$(x_n)$$ is bounded, there exists a real number $$M$$ such that all $$x_n\leq M$$. So then $$\left|\dfrac{x_n}{n^k} -0\right|\leq M\dfrac{1}{n^{k}}$$.

From here, I know that I am looking to be able to conclude that $$\left|\dfrac{x_n}{n^k} -0\right|\leq M\dfrac{1}{n^{k}} < M\dfrac{ε}{M} = ε$$, but I am not sure how to justifiably draw this conclusion.

2 Answers

All you need is $$n >M^{1/k}$$. If $$m$$ is the largest integer not exceeding $$M^{1/k}$$ then you get $$|\frac {X_n} {n^{k}} -0| < \epsilon$$ for all $$n >m$$.

$$|\dfrac{x_n}{n^k}| \le M \dfrac{1}{n^k} \le \dfrac{M}{n},$$ for $$k \in \mathbb{Z^+}.$$

Let $$\epsilon >0$$ be given.

Archimedean principle:

There is a $$n_0 \in \mathbb{Z^+}$$ with

$$n_0 > \dfrac{M}{\epsilon}$$

For $$n \ge n_0$$:

$$|\dfrac{x_n}{n^k}| \le \dfrac{M}{n} \le \dfrac{M}{n_0} < \epsilon.$$