Triple the sequence!

From the first equation

$b_n + c_n = 2n + 1 - a_n$

From third equation

$b_n.c_n = \dfrac{-1}{a_n}$

Now simplifing the second equation

$a_n(b_n + c_n) + b_n.c_n = 2n -1$

Substituting the values we will get a cubic equation

$\displaystyle a_n^{3} - (2n+1)a_n^{2} + (2n-1)a_n + 1 = 0$

By trial and error method one root of the equation is 1, the other two roots can be found easily by Dividing cubic by $a_n -1$ to get the quadratic

$a_n^{2} - 2n.a_n - 1 = 0$

(Note:-these roots of the equation are $a_n, b_n, c_n$ and we have to choose the minimum root.)

The Roots of the equation are $a_n = n \pm \sqrt{n^{2} +1}$

Minimum root is $n - \sqrt{n^{2} + 1}$

$\therefore$

$\displaystyle \lim_{n \to \infty } n \times a_n = \lim_{n \to \infty } n(n - \sqrt{n^{2} +1})$

Taking $n^{2}$ common from root

$\displaystyle \lim_{n \to \infty } n^{2} - n^{2}(\sqrt{1 + \frac{1}{n^{2}}})$

We know,

$(1 + x)^{n} = 1 + nx$ when x<<1

So our limit reduces to

$\displaystyle \lim_{n \to \infty } n^{2} - n^{2}( 1 + \dfrac{1}{2n^{2}}) = \boxed{\boxed{\dfrac{-1}{2}}}$