I'm having some difficulties with the following problem:
Let $n\in \mathbb N$ and: $$ x_n = \frac{n-3}{\sqrt{n^2+1}} $$ Prove $x_n$ is a monotonic sequence starting from some $n_0$.
I've applied the following test to prove that. Suppose $\frac{x_n}{x_{n+1}} > 1$
$$ \frac{n-3}{\sqrt{n^2+1}}\cdot\frac{\sqrt{(n+1)^2+1}}{n-2} > 1 \iff \\ \iff \frac{n-3}{n-2} >\frac{\sqrt{n^2+1}}{\sqrt{(n+1)^2+1}} $$
Now squaring both sides (this is BTW where i think things get wrong): $$ \left(\frac{n-3}{n-2}\right)^2 >\frac{{n^2+1}}{{(n+1)^2+1}} \iff \\ \iff (n-3)^2((n+1)^2+1)>(n^2 + 1)(n-2)^2 $$
After some algebraic transformations one may get: $$ 3 n^2 - 5n - 7 < 0 $$
Now find $n_0 \ge 1$ starting from which the inequality holds: $$ n_{1,2} = \frac{5 \pm\sqrt{109}}{6} $$
So from this $n \approx2.57 < 3$. So for $n \ge 3$ the inequality holds. But that's not true! From the graph it's crystal clear that $x_n$ is monotonically increasing $\forall n \ge 1$.
Where did things go wrong? I believe it's a consequence of squaring the inequality, but can't see how it went wrong.
For all $n\geq3$ we obtain: $$x_{n+1}-x_n=\frac{n-2}{\sqrt{n^2+2n+2}}-\frac{n-3}{\sqrt{n^2+1}}=$$ $$=\frac{2(3n^2-5n-7)}{\sqrt{(n^2+2n+2)(n^2+1)}((n-2)\sqrt{n^2+1}+(n-3)\sqrt{n^2+2n+2})}>0,$$ which says that our sequence increases for $n\geq3$.