If $\lvert x_n \lvert \le \frac{2n^2+3}{n^3+5n^2+3n+1}$ then $\{ x_n \}$ is Cauchy

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If $\lvert x_n \lvert$ is a sequence that satisfies $\lvert x_n \lvert \le \frac{2n^2+3}{n^3+5n^2+3n+1}$, then $\{ x_n \}$ is Cauchy.

I think that $\frac{2n^2+3}{n^3+5n^2+3n+1}$ is increasing with the lower bound of $\frac{1}{2}$. However, what if $x_n = \{\frac{1}{2}, -\frac{1}{2}, ...\}$, an alternating sequence satisfying this condition?

Can please someone point out where's my error?

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First, you want to take the limit of this sequence. You can do that using l'Hopital's rule, or you could try multiplying by $$\frac{n^{-3}}{n^{-3}}.$$ If you do that, you'll notice many of the terms get small in the limit, and you can easily find the limit. (It's not $1/2$, actually.)

If the limit is zero, then the claim is true by the sandwich theorem. Otherwise, the counterexample you described (or a modified version) will work!