The Cauchy Sequence is that: $\forall \epsilon > 0 \,\,\exists n_0 \in \mathbb{N}$ such that $\forall m>n \ge n_0 \,\,|x_m - x_n| \le \epsilon$. Let's negate it, we will get ∃>0 ∀0∈ℕ such that ∃>≥0 |−|≥.
I have tried to estimate it from above, but could not take it to the end.
Well, we can feel that $x_{2k} \to 1$ while $x_{2k + 1} \to -1$ so for a large enough $N$ if we have $n > N; n$ even and $m > N; m$ odd we should have
$|x_n - x_m| \approx 2> \epsilon$ for a small epsilon.
Let's get that in detail.
Let $\epsilon > 0$. Suppose $2k, 2k+ 1 > N$. with
$|x_{2k} - x_{2k+1}| < \epsilon$.
That means $|\frac {(2k)^2+3}{(2k)^2+ 5(2k) + 1} -(-\frac {(2k+1)^2+3}{(2k+1)^2+ 5(2k+1) + 1})| =$
$|\frac {4k^2 + 3}{4k^2 + 10k + 1} + \frac {4k^2 + 4k +4}{4k^2 + 14k + 7}|< \epsilon$.
But $\frac {4k^2 + 4k + 4}{4k^2 + 14k +7} > \frac {4k^2 +3}{4k^2 + 14k +7}$ and $\frac {4k^2 + 3}{4k^2 + 10k + 1} > \frac {4k^2 +3}{4k^2 + 14k +7}$ so (assuming $k > \frac {N}2 > 0$) we have
$2\frac {4k^2 +3}{4k^2 + 14k +7} < |\frac {4k^2 + 3}{4k^2 + 10k + 1} + \frac {4k^2 + 4k +4}{4k^2 + 14k + 7}|< \epsilon$
So $\epsilon$ must be larger than $2\frac {4k^2 +3}{4k^2 + 14k +7}=2(1 - \frac {14k+4}{4k^2+14k + 7})> 2(1-\frac {14k}{4k^2}) = 2- \frac {14}{k}$.
So if $k > 14$ then $\epsilon > 1$ .
so for any $\epsilon < 1$ we can't have an $N$ where $n, m > N$ imply $|x_n - x_m| < \epsilon$ as taking any $2k, 2k+1 > \max(N, 28)$ we have $|x_n - x_m| > \epsilon$.
SO not cauchy.