Prove that if $\{x_n\}$ is a sequence that satifies the given condition then it is cuachy

Question

$$\vert x_n \vert \leq \frac{2n^2+}{n^3+5n^2+3n +1}$$

By dividing all the terms in the numerator by $n^2$ and by dividing all the terms in the denominator by $n^3$ and taking the limit I arrive at:

$$-2 \leq \vert x_n \vert \leq 2$$

Therefor $$\vert x_n \vert$$ is bounded by 2.Where do I go from here? does this even help? The definition in the book is that A sequence of points $x_n \in \mathbb{R}$ is said to be cauchy if and only if for every $\epsilon \gt 0$ there is an $n \in \mathbb{N}$ such that $n,m$ imply $\vert x_n-x_m \vert \leq \epsilon$

Answer 1

Since$$(\forall n\in\mathbb N):0\leqslant\lvert x_n\rvert\leqslant\frac{2n^2}{n^3+5n^2+3n+1}$$and since$$\lim_{n\to\infty}\frac{2n^2}{n^3+5n^2+3n+1}=0,$$it follows from the sequeeze theorem that $\lim_{n\to\infty}x_n=0$. And every convergent sequence is a Cauchy sequence.

Answer 2

We need to factor out $n^2$ both from numerator and denominator, to obtain

$$\vert x_n \vert \leq \frac{2n^2}{n^3+5n^2+3n +1}=\frac{n^2}{n^2}\frac{2}{n+5+3/n +1/n^2}\to 1\cdot 0=0$$

Answer 3

Let $n \ge 1$;

$|x_n|\le \dfrac{2n^2}{n^3+5n^2+3n+1} \le \dfrac{2n^2}{n^3}=\dfrac{2}{n}.$

Let $\epsilon$ be given.

Choose $n_0$ s.t. $n_0 > 4/\epsilon$. (Archimedean principle)

For $n\ge m \ge n_0$ we have

$|x_n-x_m| \le |x_n| +|x_m|$

$ \le 2/n +2/m < 4/n_0 <\epsilon$.