Prove $ x_n<\frac{n+1}2$ by induction.

if $ \ x_{n}\leq\frac{n}{2} $ then$ \ x_{n+1}=x_{n}+\frac{x_{n}^{2}}{n^2} \leq \frac{n}{2}+\frac{1}{n^2}\frac{n^2}{4}\leq\frac{n+1}{2}$ now see $ \ x_{1}\leq\frac{1}{2} $ so by induction $ \ x_{n}\leq\frac{n}{2} $ so $ \ x_{2001}\leq1000.5<1001 $ done

http://www.artofproblemsolving.com/blog/42821

Obviosly $x_n$ is monotonically increase. If $x_n<C$ and $(C+\delta )^2(\frac{1}{n^2}+...+\frac{1}{(n+m-1)^2})<\delta$, then $x_{n+m}<C+\delta.$ Because $\frac{1}{k^2}<\int_{k-0.5}^{k+0.5}\frac{dx}{x^2}$, we get if $x_n<C$ and \[(C+\delta)^2(\frac{1}{n-0.5}-\frac{1}{n+m-0.5})=(C+\delta)^2\frac{m}{(n-0.5)(n+m-0.5)}<\delta\] then $x_{n+m}<C+\delta.$ It gives, that $x_n$ is bounded. We can calculate $x_4=\frac{795}{768}$. Take $C+\delta =1.5$ it gives $x_{13}<1.5$. Then take $\delta =\frac{19-\sqrt{325}}{4}<0.25$. It gives $x_n<C+\delta \ \forall n$. We prove $x_n<1.75 \ \forall n$.

In fact, we can prove that ${{x}_{n}}<\frac{3}{2}$ is right for any $n\in {{\mathbf{N}}^{+}}$.