convergence divergence - Show that the sequence given by $x_{n+1}=x_n+frac{sqrt {|x_n|}}{n^2}$ is convergent

My Try:

It is clear that $x_n$ is monotonically increasing. If we assume that the sequence converges to $a$ then $\displaystyle a=a+\frac{\sqrt{|a|}}{n^2}$. Hence $a=0$. So, I was going to prove that the sup of the sequence is $0$. But failed. Can somebody please help me to complete the proof

Answer

$$x_{n+1} = (\sqrt{x_n}+\frac1{2n^2})^2-\frac1{4n^4} \lt (\sqrt{x_n}+\frac1{2n^2})^2$$
so
$$\sqrt{x_{n+1}} \lt \sqrt{x_n} + \frac1{2n^2}$$
hence
$$\sqrt{x_{n+1}}-\sqrt{x_1} \lt \frac12\sum_{k=1}^n \frac1{n^2} \lt \frac{\pi^2}{12}$$