<s>[INST] How to prove that $ \frac{n+1}{4n^2+3}$ is a Cauchy sequence? By definition, I need to show that for any $\epsilon \gt 0$ there exists $N \in \mathbb{N}$ such that for any $m,n \gt N$:
$ \lvert a_n - a_m \rvert \lt \epsilon$
So I write,
$ |\frac{n+1}{4n^2+3} - \frac{m+1}{4m^2+3} |\leq | \frac{n+1}{4n^2+3}| + |\frac{m+1}{4m^2+3}| = \frac{n+1}{4n^2+3} + \frac{m+1}{4m^2+3} \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
From here I try to show,
$\frac{n+1}{4n^2+3} \lt \frac{\epsilon}{2} \quad \quad \Rightarrow \quad \quad  4\epsilon n^2 -2n +3\epsilon -2 \gt 0$
$n \gt \frac{1}{4\epsilon} + \frac{1}{4\epsilon} \sqrt{1-12\epsilon^2+8\epsilon}$
Can someone please explain to me what should I do from here?
Am I in the right direction at all?
 [/INST] $4n^2+3\geq 4n^2$. Therefore, $\frac{n+1}{4n^2+3} \leq \frac{2n}{4n^2}= \frac{1}{2n}$
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