Problem

x_1 = \sqrt{3}

and

x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n}}

\forall n \in N

, then

\lim_{n \to \infty} 2^n x_n

equal to a.

\frac{3}{2\pi}

\frac{2}{3\pi}

\frac{2\pi}{3}

\frac{3\pi}{2}