Convergence of a sequence with $(x_n^2)$ is non-increasing. Let $(x_n)$ a real sequence such that $(x_n^2)$ is non-increasing and $(x_{n+1}-x_n)$ converges to $0$. Prove that $(x_n)$ converges.
My attempt:
$(x_n)^2$ is a decreasing sequence and $(x_n^2)\geq 0$ therefore the sequence converges to $l\geq 0$
If $l=0$ therefore $(x_n)$ converges to $0$.
I haven't managed to do the case $l\neq 0$, I tried with $\epsilon$ but I didn't succeed.
Thank you in advance for your help,
 A: The bounded and monoton sequence $|x_n|$ must converge to some limit $l$.
Let $M=\{\,n\in\mathbb N\mid x_n>0, x_{n+1}\le 0\,\}$.
If $M$ is finite, almost all $x_n$ have the same sign, which implies $x_n\to l$ or $x_n\to-l$ depending on that sign.
If on the other hand $M$ is infinite, we have
$$\lim_{M\ni n\to \infty} x_n=l $$
$$\lim_{M\ni n\to \infty} x_{n+1}=-l $$
$$ \lim_{M\ni n\to \infty} (x_{n+1}-x_n)=0 $$
hence $l=0$. But then $|x_n|\to 0$ implies $x_n\to 0$.
A: If $x_n^2 \downarrow l>0$, then  for every $n$ we have either $x_n\ge \sqrt{l}$ or $x_n \le -\sqrt{l}$. Now, from $x_{n+1}-x_{n} \to 0$ it follows that there exists $N$ such that for every $n>N$  $x_{n+1}$  and $x_n$ must have the same sign. 
Therefore all $x_n$ can be assumed positive for $n>N$. But in this case from $x^2_{n+1} \ge x_n^2$  it follows that $x_{n+1} \ge x_n$ $n> N$. 
Therefore $x_n$ is non increasing and bounded from below by $\sqrt{l}$. Therefore it is convergent. The same is the case if we assume all $x_n<-\sqrt{l}$ for $n>N.$
