Analysis problem from Romanian Contest - 2 sequences which forms another one Let $a,b$ be 2 real numbers, and the sequences $(a_n)_{n \geq 1}, (b_n)_{n \geq 1}$ defined by $a_{1}=a$, $b_{1}=b$, $a^2+b^2 <1$ and 
\begin{cases}
a_{n+1}=\frac{1}{2}\left(a_{n}^{2}-\frac{b_{n}^{2}}{n^{2}}\right), \mbox{ }(\forall) n \geq 1\\
b_{n+1}=-\left(1+\frac{1}{n}\right)a_{n}b_{n}, \mbox{ }(\forall) n \geq 1.
\end{cases}
Prove that the sequence $\displaystyle x_{n}=\frac{a_{n}\cdot b_{n}}{n}, (\forall) n \geq 1$ is convergent and calculate $\lim\limits_{n \to \infty}{x_{n}}.$
Seems hard... Thanks for your help! 

i try to find the regular of this sequence :
$a_{2}=\frac{1}{2}$$(a^2-b^2)$$,$$b_{2}=-2ab$
$a_{3}=\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4})$$,$$b_{3}=(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{3}{2})$
$a_{4}=\frac{1}{2}((\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4}))^{2}$$-(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{3}{2})^{2}\cdot\frac{1}{9})$$,$
$b_{4}=\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4})$$\cdot$$(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{4}{3})\cdot(-\frac{3}{2})$
$......$
it seems that $b_{n}$ is always less than $1$
then, $limx_{n}=lim\frac{a_{n}\cdot b_{n}}{n}$$\longrightarrow$$\frac{b_{n+1}}{n}$
if $b_{n}\le1$ , it is obviously that $x_{n}$ is convergent, but i am not sure about how to use mathematical deduction to prove $b_{n}$ is always less than $1$ ?
i guess that, the hint is :
$ab\ge-b^{2}-a_{2}-b_{2}$
$a_{2}b_{2}\ge-b_{2}^{2}/4-a_{3}-b_{3}$
$a_{3}b_{3}\ge-b_{3}^{2}/9-a_{4}-b_{4}$
$......$
here i use the deduce method to find :
$a_{n}b_{n}\le\frac{1}{2^{2n-1}}$ $,$   $b_{n}\le{\frac{n}{2^{n}}}$$\Longrightarrow$$a_{n+1}b_{n+1}\le\frac{1}{2^{2n+1}}$ $,$   $b_{n+1}\le{\frac{n+1}{2^{n+1}}}$
$\Longrightarrow$$a_{n}+b_{n}\le{a_{n-1}+b_{n-1}}\le{a_{n-2}+b_{n-2}}\le......\le1$$\Longrightarrow$$b_{n}^{2}\le{b_{n-1}^{2}}\le{b_{n-2}^{2}}\le......\le{b^{2}}$
since,  $\frac{n+1}{2^{n}}<1$
is it helpful? thank you !
