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
 A: If your argument were correct it would show that the sequence diverges if $x_1>0$, because in that case the sup certainly can't be $0$. But your argument isn't correct. If you assume that $x_n\to a$ and plug that into the recurrence you get just $a=a$, which doesn't tell you much about $a$. (You forgot that $n\to\infty$ as $n\to\infty$.)
It's true that the sequence is increasing. You're not asked to find the limit. So you just have to show the sequence is bounded.
If $x_n<1$ for all $n$ then the sequence is bounded. Suppose not: There exists $N$ with $x_N\ge1$. Then $\sqrt{x_N}\le x_N$, so $$x_{N+1}\le\left(1+\frac1{N^2}\right)x_N.$$
Similarly for $x_{N+2}$ you get $$x_{N+2}\le\left(1+\frac1{N^2}\right)\left(1+\frac1{(N+1)^2}\right)x_N.$$And so on. You only need to show that $$\prod_{n=N}^\infty\left(1+\frac1{n^2}\right)<\infty.$$Read a little about infinite products, or take the logarithm, and you see this follows from $\sum\frac1{n^2}<\infty$.
A: 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 $|x_j| \le M^2$.
A: $$
x_{n+1} = (\sqrt{x_n}+\frac1{2n^2})^2-\frac1{4n^4} \lt (\sqrt{x_n}+\frac1{2n^2})^2
$$
so
$$
\sqrt{x_{n+1}} \lt \sqrt{x_n} + \frac1{2n^2}
$$
hence
$$
\sqrt{x_{n+1}}-\sqrt{x_1} \lt \frac12\sum_{k=1}^n \frac1{n^2} \lt \frac{\pi^2}{12}
$$
