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

No, the limit is not $0$ in general.  
Hint: 
$$x_n - x_1 = \sum_{j=1}^{n-1} (x_{j+1} - x_j) = \sum_{j=1}^{n-1} \dfrac{\sqrt{|x_j|}}{j^2} $$
Find $M$ such that $|x_1| < M^2$ and $M \sum_{j=1}^\infty 1/j^2 < M^2 - x_1$, and show that
all $|x_j| \le M^2$.
