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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.

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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 $n >m$.

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$|\dfrac{x_n}{n^k}| \le M \dfrac{1}{n^k} \le \dfrac{M}{n},$ 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 $:

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

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